Sequence and Series, Class 11 Mathematics R.D Sharma Solutions

ARITHMETIC PROGRESSION

Page 19.4 Ex 19.1

Q1.

Answer :

Given : an = n2-n+1For n=1, a1 = 12-1+1 = 1For n=2, a2 = 22-2+1 = 3For n=3, a3 = 32-3+1 = 7For n=4, a4 = 42-4+1 =13For n=5, a5 = 52-5+1 = 21
Thus, the first five terms of the sequence are 1, 3, 7, 13, 21.

Q2.

Answer :

Given:
an = n³ − 6n² + 11n − 6, n ϵ N
For n =1, a1 = 13-6×12+11×1-6 = 0For n = 2, a2 = 23-6×22+11×2-6 = 0For n =3, a3 = 33-6×32+11×3-6 = 0For n =4, a4 = 43-6×42+11×4-6 =6>0For n =5, a5 = 53-6×52+11×5-6 =24 >0and so onThus, the first three terms are zero and the rest of the terms are positive in the sequence.

Q3.

Answer :

(i) We have: -1-3 = -4,-5-(-1) = -4,-9-(-5) = -4…Thus, the sequence is an A.P. with the common difference being -4.The next three terms are as follows:-9-4 = -13-13-4 = -17-17-4 = -21(ii) We have:1/4-(-1) = 5/43/2-1/4 = 5/411/4-3/2 = 5/4Thus, the sequence is an A.P. with the common difference being (5/4).The next three terms are as follows:11/4+5/4 = 16/4 = 416/4+5/4 = 21/421/4+5/4 = 26/4

(iii) We have: 32 – 2 = 2252 – 32 = 2272 – 52 = 22Thus, the sequence is an A.P. with the common difference being (22).The next three terms are as follows:72+22 = 9292+22 = 112112+22 = 132(iv) We have: 7-9 = -25-7 = -23-5 = -2Thus, the sequence is an A.P. with the common difference being (-2).The next three terms are as follows:3-2 = 11-2 = -1-1-2 = -3

Q4.

Answer :

an = 2n+7∴ a1 = 2×1+7 = 9 a2 =2×2+7 =11 a3 =2×3+7 = 13a4 = 2×4+7 = 15and so onSo, common differenced=11-9=2Thus, the above sequence is an A.P. with the common difference as 2a7 = 2×7+7 = 21

Q5.

Answer :

We have:
an = 2n² + n + 1
a1 = 2×12+1+1 = 4,a2 = 2×22+2+1 = 11a3 = 2×32+3+1 = 22 a2-a1= 11-4 = 7 and a3-a2= 22-11 = 11Since, a2-a1≠a3-a2Hence,it is not an AP.

Q6.

Answer :

Given:
a₁ = 3
And, a_n = 3a_n-1 + 2 for all n > 1

a2 = 3a2-1+2 = 3a1+2 = 11a3 = 3a3-1+2 = 3a2+2 = 35a4 = 3a4-1+2 = 3a3+2 = 107Thus, the first four terms of the sequence are 3, 11, 35, 107.
Q7.

Answer :

(i) a₁ = 1, a_n = a_n-1 + 2, n ≥ 2
a2 = a1+2 = 1+2 = 3a3 = a2+2 = 5a4 = a3+2 = 7a5 = a4+2 = 9

Hence, the five terms are 1, 3, 5, 7 and 9.

(ii) a₁ = 1 = a₂, a_n = a_n-1 + an − 2, n > 2
a3 =a2+a1 = 1+1=2a4 = a3+a2 =2+1 = 3a5 = a4+a3 = 3+2 = 5

Hence, the five terms are 1, 1, 2, 3 and 5.

(iii) a1 = a2 = 2, an = an − 1 − 1, n > 2
a3 = a2-1 = 2-1= 1a4 = a3-1 = 1-1= 0a5 = a4-1 = 0-1= -1

Hence, the five terms are 2, 2, 1, 0 and -1.

Q8.

Answer :

a₁ = 1 = a₂, a_n = a_n-1 + a_n-2 for n > 2
Then, we have:
a3 = a2+a1 = 1+1 = 2a4 = a3+a2 = 2+1 = 3a5 =a4+a3 = 3+2 = 5a6 = a5+a4 = 5+3 = 8For n=1, an+1an = a2a1 = 11=1For n=2, an+1an = a3a2 = 21=2For n=3, an+1an = a4a3 = 32For n=4, an+1an = a5a4 = 53For n=5, an+1an = a6a5 = 85

Page 19.12 Ex 19.2

Q1.

Answer :

(i) 1, 4, 7, 10…

We have:
a = 1
d = 4-1 = 3

a10 =a+(10-1)d an=a+n-1d = a +9d = 1+9×3 = 28

(ii) 2, 32, 52…
We have:
a = 2d = 32 – 2 = 22

a18 = a+18-1d an=a+n-1d = a+17d = 2+1722 = 2+342 = 35 2

(iii) 13, 8, 3, −2…
We have:
a = 13d = 8 -13 = -5

an = a+(n-1)d = 13+(n-1)-5 = 13-5n+5 = 18-5n

Q2.

Answer :

Let the sequence < an > be an A.P. with the first term being A and the common difference being D.
To prove: am +n +am − n = 2am

LHS: am +n +am − n

= A+(m+n-1)D + A+(m-n-1)D {∵an = a+(n-1)d}= A+mD+nD-D +A+mD-nD-D=2A+2mD-2D …(i)

RHS: 2am

= 2[A+(m-1)D]= 2A +2mD-2D …(ii)

From (i) and (ii), we get:
LHS = RHS
Hence, proved.

Q3.

Answer :

(i) 3, 8, 13…
Here, we have:
a = 3
d = 8-3 =5
Let an = 248⇒ a+n-1d = 248⇒3+n-15 = 248⇒n-15 =245⇒n-1 = 49⇒ n= 50

Hence, 248 is the 50th term of the given A.P.

(ii) 84, 80, 76…
Here, we have:
a = 84
d = 80-84 = -4
Let an =0⇒a+(n-1)d = 0⇒84 + (n-1)-4 = 0⇒ (n-1)-4 = -84⇒(n-1) = 21⇒ n = 22

Hence, 0 is the 22nd term of the given A.P.

(iii) 4, 9, 14…
Here, we have:
a = 4
d = 9-4 = 5
Let an = 254⇒a+n-1 d = 254⇒4+n-1 5 = 254⇒n-1 5 =250⇒n-1 = 50⇒ n= 51

Hence, 254 is the 51st term of the given A.P.

Q4.

Answer :

(i) 7, 10, 13…
Here, we have:
a = 7
d = 10-7 = 3Let an = 68⇒ a+(n-1) d = 68⇒7+(n-1)(3) = 68⇒(n-1)(3) = 61⇒(n-1) = 613⇒n = 613 +1 = 643

Since n is not a natural number.So, 68 is not a term of the given A.P.

(ii) 3, 8, 13…
Here, we have:
a = 3
d = 8-3=5Let an = 302⇒ a+n-1d = 302⇒3+n-15 = 302⇒n-15 =299⇒n-1 = 2995⇒n = 2995+1 = 3045
Since n is not a natural number.So, 302 is not a term of the given A.P.

Q5.

Answer :

(i) 24, 2314, 2212, 2134…
This is an A.P.
Here, we have:
a = 24
d = 2314- 24 = -34Let the first negative term be an.Then, we have: an <0⇒a+n-1 d<0⇒24+n-1 -34<0⇒24-3n4+34<0⇒24+34<3n4⇒994<3n4⇒99<3n⇒n >33

Thus, the 34th term is the first negative term of the given A.P.

(ii)
(a) 12 + 8i, 11 + 6i, 10 + 4i…
This is an A.P.
Here, we have:
a = 12 + 8i
d = 11+6i – 12-8i = -1-2iLet the real term be an = a+n-1d. an = 12+8i+n-1-1-2i = 12+8i+ -n+1-2in+2i = 12+8i-n+1-2in+2i = 13-n +8-2n+2i = 13-n +10-2nian has to be real.∴10-2n = 0⇒ n= 5

(b) 12 + 8i, 11 + 6i, 10 + 4i…
This is an A.P.
Here, we have:
a = 12 + 8i
d = 11+6i – 12-8i = -1-2iLet the imaginary term be an = a+n-1d an = 12+8i+n-1-1-2i = 12+8i+ -n+1-2in+2i = 12+8i-n+1-2in+2i = 13-n +8-2n+2i = 13-n +10-2nian has to be imaginary.∴ 13-n = 0⇒ n=13

Q6.

Answer :

(i) 7, 10, 13…43
Here, we have:
a = 7
d = (10-7) = 3an = 43
Let there be n terms in the given A.P.

Also, an =a+n-1d⇒43 = 7+n-13⇒36 = n-13⇒12 = n-1⇒13 = n

Thus, there are 13 terms in the given A.P.

(ii) -1,-56,-23,-12, …,103

Here, we have:
a = -1
d = -56–1 = 1-56= 16an = 103
Let there be n terms in the given A.P.

Also, an =a+n-1d⇒103= -1+n-116⇒133 = n-116⇒26= n-1⇒27 = n

Thus, there are 27 terms in the given A.P.

Q7.

Answer :

Here, a = 5, d = 3, a_n = 80
Let the number of terms be n.
Then, we have:

an = a+n-1d⇒80 = 5+n-13⇒75 = n-13⇒25 = n-1⇒26 = n

Thus, there are 26 terms in the given A.P.

Q8.

Answer :

Given:

a6 = 19⇒ a+6-1d = 19 an=a+n-1d⇒ a+5d=19 ..1And, a17 = 41⇒a+17-1d=41 an=a+n-1d⇒a+16d = 41 ..2Solving the two equations, we get,16d-5d = 41-19⇒11d = 22⇒d = 2 Putting d = 2 in the eqn 1, we get:a+5×2 = 19⇒a = 19-10⇒a= 9
We know:
a40 =a+40-1d an=a+n-1d = a+39d = 9+39×2 = 9+78 = 87

Q9.

Answer :

Given:
a9 = 0 ⇒a+9-1d=0 an=a+n-1d⇒a+8d = 0⇒a = -8d …(i)

To prove:
a29 = 2a19

Proof:
LHS: a29 =a+29-1d= a+28d=-8d+28d From(i)= 20d

RHS: 2a19=2a+19-1d= 2(a+18d)= 2a+36d= 2(-8d)+36d From(i)= -16d+36d= 20d

LHS = RHS
Hence, proved.

Q10.

Answer :

Given:
10a10 = 15a15⇒10a+10-1d=15a+15-1d⇒10(a+9d) = 15(a+14d)⇒10a+90d = 15a+210d⇒0 = 5a+120d⇒0 = a+24d⇒a = -24d …(i)

To show:
a25 = 0⇒LHS: a25=a+25-1d =a+24d=-24d +24d From(i)= 0 = RHSHence, proved.

Q11.

Answer :

Given:

a10 = 41⇒a+10-1d=41 an=a+n-1d⇒a+9d = 41 And, a18=73⇒a+18-1d=73 an=a+n-1d⇒a+17d =73 Solving the two equations, we get:⇒17d-9d = 73-41⇒8d = 32⇒d = 4 …(i)Putting the value in first equation, we get:a+9×4 = 41⇒a+36 = 41⇒a = 5 …(ii)

a26 = a+26-1d an=a+n-1d⇒ a26 = a+25d ⇒a26 =5+25×4 From (i) and (ii)⇒a26 =5+100 = 105

Q12.

Answer :

Given:
a24 = 2a10⇒a+24-1d=2a+10-1d⇒a+23d = 2(a+9d)⇒a+23d = 2a+18d⇒5d = a …(i)

To prove:a72 = 2a34LHS: a72 =a+72-1d⇒a72 =a+71d⇒a72 =5d+71d From(i)⇒a72 =76d

RHS: 2a34 = 2a+34-1d⇒2a34 =2 a+33d⇒2a34 =2(5d+33d) Form(i)⇒2a34 = 238d⇒2a34 =76d
∴ RHS = LHS
Hence, proved.

Q13.

Answer :

Given:
am+1 = 2an+1⇒a+(m+1-1)d = 2[a+(n+1-1)d]⇒a+md = 2(a+nd)⇒a+md = 2a+2nd⇒0 = a+2nd-md ⇒nd=md-a2 …(i)

To prove:

a3m+1 = 2am+n+1LHS: a3m+1 =a+(3m+1-1)d⇒a3m+1 =a+3mdRHS: 2am+n+1 = 2[a+(m+n+1-1)d]⇒2am+n+1 =2(a+md+nd)⇒2am+n+1 = 2a+md+md-a2 From(i)⇒2am+n+1 = 22a+2md+md-a2

⇒2am+n+1=2a+3md2⇒2am+n+1=a+3md

∴ LHS = RHS

Q14.

Answer :

Given:
nth term of the A.P. 9, 7, 5… is the same as the nth term of the A.P. 15, 12, 9…

Considering 9, 7, 5…a = 9, d= 7-9 = -2nth term = 9+(n-1)(-2) an=a+n-1d = 9-2n+2 = 11-2n …(i)

Considering 15, 12, 9, …a = 15, d = 12-15 = -3nth term =15+(n-1)(-3) an=a+n-1d = 15-3n+3 = 18-3n …(ii)

Equating (i) and (ii), we get:
11-2n = 18-3n⇒n = 7

Thus, 7th terms of both the A.P.s are the same.

Page 19.13 Ex 19.2

Q15.

Answer :

Let the three terms of the A.P. be a-d, a, a+d.Then, we have:a-d+a+a+d = 21⇒3a =21⇒a = 7 ….(i)Also, (a-d)(a+d) – a = 6⇒a2-d2 -a = 6⇒49-d2 -7 = 6⇒36=d2⇒±6 = dWhen d=6, a=7, we get:1, 7, 13 When d=-6, a=7, we get:13, 7, 1

Q16.

Answer :

Given:
Let the first term of the A.P. be a and the common difference be d.
a4=3a⇒ a+4-1d = 3a⇒ a+3d = 3a⇒3d = 2a⇒a=3d2 …(i)And, a7-2a3=1⇒a+7-1d-2a+3-1d=1⇒a+6d -2(a+2d) =1⇒a+6d-2a-4d =1⇒-a+2d = 1⇒-3d2+2d = 1 From (i) ⇒-3d+4d2=1⇒d2=1⇒d=2
Putting the value in (i), we get:a=3×22⇒a = 3

Q17.

Answer :

Given:
a6 =12⇒a+6-1d=12⇒a+5d = 12 …(i)a8 =22⇒a+8-1d=22⇒a+7d = 22 …(ii)Solving (i) and (ii), we get:2d = 10⇒d = 5Putting the value of d in (i), we get: a+5×5 = 12⇒a=12-25 = -13∴ a2 = a+2-1d=a+d = -13+5 = -8Also, an = a+(n-1)d =-13+(n-1)5 = -13+5 n-5 = 5n-18

Q18.

Answer :

The two digit numbers that are divisible by 3 are:
12, 15, 18…96, 99
This is an A.P. whose first term is 12 and the common difference is 3.
We have:an = 99⇒12+(n-1)3 = 99⇒(n-1)3 =87⇒(n-1) = 29⇒n = 30
Thus, there are 30 such terms.

Q19.

Answer :

Given:
a = 7, n= 60, l= 125
l = a+(n-1)d⇒125 =7+(60-1)d⇒125 = 7+59d⇒118 = 59d⇒2 = d

a32 = a+32-1d =a+31d = 7+31×2 = 7+62 = 69

Q20.

Answer :

Let a be the first term and d be the common difference. Then,
a4+a8 = 24⇒a+4-1d+a+8-1d=24⇒a+3d+a+7d = 24⇒2a+10d = 24 ⇒a+5d = 12 …(i)Also, a6+a10 =34⇒a+6-1d+a+10-1d=34⇒a+5d+a+9d = 34⇒2a+14d = 34⇒a+7d = 17 …(ii)Solving (i) and (ii), we get:2d = 5⇒d = 52Substituing the value in (i), we get:a+552 = 12⇒a+252 = 12⇒a = -12

Q21.

Answer :

Given:
First term = a
Last term = l
nth term from the beginning = a+(n-1)d, where d is the common difference.
nth term from the end = l+(n-1)(-d) =l-dn+d
Their sum = a+(n-1)d +l-dn+d
= a+nd-d+l-nd+d = a+l
Hence, proved.

Q22.

Answer :

Given:
< a_n > is an A.P.
a4a7 = 23⇒a+4-1da+7-1d= 23 ⇒a+3da+6d= 23⇒3(a+3d) = 2(a+6d) ⇒3a+9d = 2a+12d⇒a = 3d ….(i)

∴ a6a8 = a+6-1da+8-1d⇒a6a8 =a+5da+7d⇒a6a8 =3d+5d3d+7d From(i)⇒a6a8 = 8d10d⇒a6a8 = 4d5d=45

Page 19.15 Ex 19.3

Q1.

Answer :

Q2.

Answer :

Let the three numbers be a-d, a, a+d.Their sum = 27⇒ a-d+a+a+d = 27⇒3a = 27⇒a=9 …(i)Product = (a-d)a(a+d) =648⇒ a(a2-d2) = 648⇒9(81-d2) =648⇒(81-d2) =72⇒d2 = 9⇒d = ±3When a=9, d=3, we have:6, 9, 12When a=9, d=-3, we have:12, 9, 6

Q3.

Answer :

Let the four numbers be a-3d, a-d, a+d, a+3d.Sum = 50⇒ a-3d+ a-d+a+d+a+3d = 50⇒4a =50⇒a = 25 2 …(i)Also, a+3d = 4(a-3d)⇒a+3d =4a-12d⇒3a = 15d⇒a = 5d⇒252×5 = d Using (i)⇒52=dSo, the terms are as follows: 25 2-3×52, 25 2-52,25 2+52, 25 2+3×52= 5, 10, 15, 20

Q4.

Answer :

Let the angles be (a-3d)°, (a-d)°, (a+d)°, (a+3d)°.
Here, d =10
So, a-30°, (a-10)°, (a+10)°, (a+30)° are the angles of a quadrilateral whose sum is 360o.

∴ (a-30)°+(a-10)°+(a+10)°+(a+30)° = 360°⇒4a = 360⇒a = 90

The angles are as follows:90-30°, (90-10)°, (90+10)°, (90+30)°, i.e. 60°, 80°, 100°, 120°

Q5.

Answer :

Let the numbers be (a-d), a, (a+d).Sum = a-d+a+a+d =12⇒3a = 12⇒a =4Also, (a-d)3+a3+(a+d)3 =288⇒a3-d3-3a2d+3ad2+a3+a3+d3+3a2d+3ad2 =288⇒3a3+6ad2 = 288⇒343+6×4×d2 = 288⇒192+24d2 =288⇒24d2 =96⇒d2 = 4⇒d = ±2When a=4, d=2, the numbers are 2, 4, 6.When a=4, d=-2, the numbers are 6, 4, 2.

Q6.

Answer :

Let the three numbers be (a-d), a, (a+d).Sum = 24⇒(a-d)+a+(a+d) =24⇒3a =24⇒a = 8 …(i)Product =a(a-d)(a+d) =440⇒a(a2-d2) = 440⇒8(64-d2) =440 Form (i)⇒(64-d2) =55⇒d2 = 9⇒d = ±3

With a = 8, d = 3, we have:5, 8, 11With a = 8, d =-3, we have:11, 8, 5

Page 19.28 Ex 19.4

Q1.

Answer :

(i) 50, 46, 42 … to 10 terms
We have: a = 50, d = 46-50 =-4n = 10Sn = n22a+(n-1)d= 1022×50+(10-1)(-4)= 5100-36 = 320

(ii) 1, 3, 5, 7 … to 12 terms
We have: a = 1, d = 3-1 =2n = 12Sn = n22a+(n-1)d= 1222×1+(12-1)(2)= 624 = 144

(iii) 3, 9/2, 6, 15/2 … to 25 terms
We have: a = 3, d = 9/2-3 =3/2n = 25Sn = n22a+(n-1)d= 2522×3+(25-1)(3/2)=252×42= 525

(iv) 41, 36, 31 … to 12 terms
We have: a = 41, d = 36-41 =-5n = 12Sn = n22a+(n-1)d= 1222×41+(12-1)(-5)= 6×27= 162

(v) a + b, a − b, a − 3b … to 22 terms
We have:First term = a+b, d = a-b-a-b =-2bn =22Sn = n22a+(n-1)d= 2222×(a+b)+(22-1)(-2b)= 112a-40b = 22a-440b

(vi) (x − y)², (x² + y²), (x + y)² … to n terms
We have: a = (x − y)2, d = x2+y2-(x − y)2 =2xySn = n22a+(n-1)d= n22(x − y)2+(n-1)(2xy)= n2×2(x − y)2+(n-1)(xy)= n(x − y)2+(n-1)(xy)

(vii) x-yx+y,3x-2yx+y,5x-3yx+y … to n terms
We have: a = x-yx+y, d =3x-2yx+y-x-yx+y=2x-yx+ySn = n22a+(n-1)d= n22x-yx+y+(n-1)2x-yx+y=n2(x+y)(2x-2y)+(2x-y)(n-1)=n2(x+y)2x-2y-2x+y+n(2x-y)=n2(x+y)n(2x-y)-y

Page 19.29 Ex 19.4

Q2.

Answer :

(i) 2 + 5 + 8 + … + 182
Here, the series is an A.P. where we have the following:
a = 2d=5-2 = 3an=182⇒2+(n-1)(3)=182⇒2+3n-3=182⇒3n-1=182⇒3n=183⇒n=61 Sn =n22a+(n-1)d⇒S61= 6122×2+61-1×3 =6122×2+60×3 = 5612

(ii) 101 + 99 + 97 + … + 47
Here, the series is an A.P. where we have the following:
a = 101d=99-101 = -2an=47⇒101+(n-1)(-2)=47⇒101-2n+2=47⇒2n-2=54⇒2n=56⇒n=28Sn =n22a+(n-1)d⇒ S28 = 2822×101+28-1×(-2) = 2822×101+27×(-2) = 2072

(iii) (a − b)² + (a² + b²) + (a + b)² + … + [(a + b)² + 6ab]
Here, the series is an A.P. where we have the following:
a = (a-b)2d=a2+b2- (a-b)2 = 2aban=[(a + b)2+6ab]⇒(a-b)2+(n-1)(2ab)= [(a + b)2+6ab] ⇒a2+b2-2ab+2abn-2ab=[a2+b2+2ab+6ab]⇒ a2+b2-4ab+2abn=a2+b2+8ab⇒ 2abn=12ab ⇒n=6 Sn =n22a+(n-1)d⇒S6= 622(a-b)2+6-1 2ab = 32(a2+b2-2ab)+10ab = 32a2+2b2-4ab+10ab = 32a2+2b2+6ab = 6a2+b2+3ab

Q3.

Answer :

The first n natural numbers are:
1, 2, 3, 4…
a = 1, d = 1, Total terms = n

Sn = n22a+(n-1)d⇒Sn = n22×1+(n-1)1⇒Sn = n22+(n-1)1⇒Sn = n2n+1

Q4.

Answer :

We have to find the sum of all the natural numbers that are divisible by 2 or 5
Required Sum = Sum of the natural numbers between 1 and 100 that are divisible by 2 + Sum of the natural numbers between 1 and 100 that are divisible by 5
− Sum of the natural numbers between 1 and 100 that are divisible by 2 and 5, i.e by 10

=2+4+6+8+…+100+5+10+15+…+100-10+20+30+…+100=5022+100+2025+100-10210+100=2550+1050-550=3050

Q5.

Answer :

The first n odd natural numbers are:
1, 3, 5, 7, 9…
a = 1, d = 2, Total terms = n

Sn = n22a+(n-1)d⇒Sn = n22×1+(n-1)2⇒Sn = n22+(n-1)2⇒Sn = n22n⇒Sn = n2

Q6.

Answer :

All the odd numbers between 100 and 200 are:
101, 103…199
Here, we have:
a = 101d = 2an = 199⇒101+(n-1)×2=199⇒2n-2=98⇒2n = 100⇒n = 50 Sn = n22a+(n-1)d⇒ S50 = 5022×101+(50-1)2⇒S50 = 25202+98
⇒S50=7500

Q7.

Answer :

The odd integers between 1 and 1000 that are divisible by 3 are:
3, 9, 15, 21…999
Here, we have:
a = 3, d = 6an= 999⇒3+(n-1)6=999⇒3+6n-6=999⇒6n=1002⇒n =167 Sn = n22a+(n-1)d⇒S167 = 16722×3+(167-1)6⇒S167 = 16721002 = 83667Hence, proved.

Q8.

Answer :

The integers between 84 and 719, which are multiples of 5 are:
85, 90…715
Here, we have:
a = 85d= 5an= 715⇒85+(n-1)5=715⇒ 5n-5=630⇒n=127 Sn = n22a+(n-1)d⇒ S127 = 12722×85+(127-1)5⇒S127 = 1272800 = 50800

Q9.

Answer :

The integers between 50 and 500 that are divisible by 7 are:
56, 63…497
Here, we have:
a = 56d =7 an = 497⇒56+(n-1)7=497⇒7n-7=441⇒7n=448⇒n=64 Sn = n22a+(n-1)d⇒ S64 = 6422×56+(64-1)7⇒ S64 = 322×56+63×7 = 17696

Q10.

Answer :

The even integers between 101 and 999 are:
102, 104…998
Here, we have:
a= 102d = 2 an = 998⇒102+(n-1)2=998⇒2n-2=896⇒2n = 898⇒n=449Sn = n22a+(n-1)d⇒S449 = 44922×102+(449-1)×2⇒S449 = 44921100⇒S449 = 246950

Q11.

Answer :

The integers between 100 and 550 that are divisible by 9 are:
108, 117…549
Here, we have:
a= 108 d = 9an = 549⇒108+(n-1)(9)=549⇒9n-9=441⇒9n=450⇒n=50Sn = n22a+(n-1)d⇒S50 = 5022×108+(50-1)×9⇒S50 = 25657 = 16425

Q12.

Answer :

The given sequence i.e., 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 +….. to 3n terms.

can be rewritten as 3 + 6 + 9 + …. to n terms + 5 + 9 + 13 + …. to n terms + 7 + 12 + 17 + …. to n terms

Clearly, all these sequence forms an A.P. having n terms with first terms 3, 5, 7 and common difference 3, 4, 5

Hence, required sum = n22×3+n-13+n22×5+n-14+n22×7+n-15

=n26+3n-3+10+4n-4+14+5n-5=n212n+18=3n2n+3

Q13.

Answer :

The sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 are:
103, 119…791
Here, we have:
a = 103
d = 16
an = 791
We know:an = a+(n-1)d⇒791 = 103+(n-1)×16⇒688 = 16n-16⇒704 = 16n⇒44 = nAlso, Sn = n2[2a+(n-1)d]⇒S44 = 442[2×103+(44-1)×16]⇒S44 = 22 [206+688]⇒S44 = 22 × 894 = 19668

Q14.

Answer :

(i) 25 + 22 + 19 + 16 + … + x = 115
Here, a = 25, d = -3, Sn = 115
We know:Sn = n22a+(n-1)d⇒115 = n22×25+(n-1)×-3⇒115×2 = n50-3n+3⇒230 = n53-3n⇒230 = 53n – 3n2⇒3n2-53n+230 = 0By quadratic formula: n = -b±b2-4ac2aSubstituting a=3, b=-53 and c=230, we get:n= 53±532-4×3×2302×3 = 466, 10⇒n = 10, as n≠466∴ an= a+(n-1)d⇒ x = 25+(10-1)(-3)⇒x =25-27 = -2

(ii) 1 + 4 + 7 + 10 + … + x = 590
Here, a = 1, d = 3,
We know:Sn = n22a+(n-1)d⇒590 = n22×1+(n-1)×3⇒590×2 = n2+3n-3⇒1180 = n3n-1⇒1180 = 3n2 – n⇒3n2-n-1180 = 0By quadratic formula: n = -b±b2-4ac2aSubstituting a=3, b=-1 and c=-1180, we get:⇒n= 1±12+4×3×11802×3 = -1186, 20⇒n = 20, as n≠-1186∴ an= x = a+(n-1)d⇒ x = 1+(20-1)(3)⇒x =1+60-3 = 58

Q15.

Answer :

Let the first term, the common difference and the sum of the first n terms of the first A.P. be a1, d1 and S1, respectively, and those of the second A.P. be a2, d2 and S2, respectively.Then, we have, S1 = n22a1+n-1d1 And, S2 = n22a2+n-1d2Given: S1S2 = n22a1+n-1d1 n22a2+n-1d2 = 7n+2n+4⇒S1S2 = 2a1+n-1d12a2+n-1d2= 7n+2n+4To find the ratio of the 5th terms of the two A.P.s, we replace n by (2×5-1 = 9) in the above equation:⇒ 2a1+9-1d12a2+9-1d2= 7×9+29+4⇒2a1+8d12a2+8d2= 7×9+29+4=6513 ⇒a1+4d1a2+4d2= 51=5:1

Q16.

Answer :

We have: a = -14 and Sn = 40 …(i)a5 = 2⇒a+5-1d = 2⇒-14+4d = 2⇒4d = 16⇒d = 4 …(ii)Also, Sn = n22a+(n-1)d⇒40 = n22-14+(n-1)×4 (From(i) and (ii))⇒80 = n-28+4n-4⇒80 = 4n2-32n⇒n2-8n-20 = 0⇒(n-10)(n+2) = 0⇒ n = 10, -2But, n cannot be negative.∴ n = 10

Q17.

Answer :

We have: S7 = 10⇒722a+(7-1)d = 10⇒722a+6d =10⇒a+3d = 107 …(i)Also, the sum of the next seven terms = S14 – S7 = 17⇒1422a+14-1d-722a+(7-1)d = 17⇒72a+13d-722a+6d =17⇒14a+91d – 7a-21d = 17⇒7a+70d = 17⇒a+10d = 177 …(ii)From (i) and (ii), we get:107 -3d = 177-10d⇒7d= 1⇒d = 17Putting the value in (i), we get:a+3d = 107⇒a+37 = 107⇒a = 1∴ a = 1, d = 17
The progression thus formed is 1, 87, 97, 107…

Q18.

Answer :

Given: a3 = 7, a7 – 3a3 = 2We have: a3 = 7⇒a+3-1d=7⇒ a+2d =7 …(i) Also, a7 – 3a3 = 2⇒a7 -21 = 2 (Given)⇒a+7-1d=23⇒a+6d = 23 …(ii)From (i) and (ii), we get:4d = 16⇒d = 4Putting the value in (i), we get: a+2(4) = 7⇒a = -1∴S20 = 2022-1+20-1(4)⇒S20 = 10-2+76⇒S20 = 1074 =740∴ a =-1, d = 4, S20 = 740

Q19.

Answer :

Given: a = 2, l = 50, Sn =442Now, Sn =442⇒n2a+l = 442⇒n22+50 = 442⇒n = 17 ∴ a17=50⇒a+(17-1)×d = 50 ⇒2+16d = 50⇒d = 3

Q20.

Answer :

Let total number of terms be 2n.
According to question, we have:

a1+a3+…+a2n-1 =24 …(1)a2+a4+…+a2n = 30 …(2)Subtracting (1) from (2), we get:d+d+…+upto n terms = 6⇒nd = 6 …(3)Given: a2n= a1+212⇒ a2n- a1 = 212⇒a+(2n-1)d -a = 212 [∵ a2n = a+(2n-1)d, a1 =a]⇒2nd-d = 212⇒2×6 – d = 212 From(3)⇒d = 32Putting the value in (3), we get:n = 4⇒2n = 8Thus, there are 8 terms in the progression.

To find the value of the first term:a2+a4+…+a2n = 30⇒(a+d) +(a+3d)+…+[a+(2n-1)d] = 30⇒n2a+d+a+(2n-1)d = 30Putting n=4 and d=32, we get: a = 32So, the series will be 112, 3, 412…

Q21.

Answer :

Sn = n2p⇒n22a+(n-1)d = n2p⇒2np = 2a+(n-1)d …(i)Sm = m2p⇒m22a+(m-1)d = m2p⇒2mp = 2a+(m-1)d …(ii)Subtracting (ii) from (i), we get:2p(n-m) = (n-m)d⇒2p = d …(iii)Substituing the value in (i), we get:nd = 2a+(n-1)d⇒nd -nd+d = 2a⇒a = d2= p from(iii) …(iv)∴ Sp = p22a+p-1d⇒Sp = p22p+p-12p⇒Sp = p22p+2p2-2p⇒Sp = p22p2⇒Sp = p3

Q22.

Answer :

Let a be the first term and d be the common difference.
a12 = -13⇒a+12-1d =-13⇒a+11d =-13 …(i)Also, S4 = 24⇒422a+(4-1)d = 24⇒22a+3d = 24⇒2a+3d = 12 …(ii) From (i) and (ii), we get:19d=-38⇒d = -2Putting the value of d in (i), we get:a+11-2=-13⇒a=9S10 = 1022×9+(10-1)-2⇒ S10 = 518+9-2 = 0

Q23.

Answer :

We have:a5 = 30⇒a+5-1d = 30⇒a+4d = 30 …(i)Also, a12 = 65⇒a+12-1d = 65⇒a+11d = 65 …..(ii)Solving (i) and (ii), we get:7d=35⇒d=5Putting the value of d in (i), we get:a+4×5=30⇒a=10∴ S20 = 2022×10+(20-1)×5⇒S20 = 102×10+(20-1)×5⇒S20 =1150

Q24.

Answer :

Let the A.P. be a, a+d, a+2d…∴ S1 = 2n+122a+(2n+1-1)d⇒S1 = 2n+122a+(2n)d⇒S1 = (2n+1)(a+nd) …(i)S2 = n+122a+(n+1-1)×2d⇒S2 = n+122a+2nd⇒S2 = (n+1)a+nd …(ii)From (i) and (ii), we get:S1S2 = 2n+1n+1Hence, proved.

Q25.

Answer :

Given:
Sn = 3n2For n=1, S1 = 3×12 = 3For n=2, S2 = 3×22 = 12For n=3, S3 = 3×32 = 27 and so on∴ S1 =a1 = 3a2 = S2-S1 = 12-3 = 9a3 = S3-S2 = 27-12 = 15and so onThus, the A.P. is 3, 9, 15…

Page 19.30 Ex 19.4

Q26.

Answer :

We have:Sn = nP+12n(n-1)QFor n =1, S1 = P+0=PFor n =2, S2 = 2P+Q Also, a1 = S1 = P,a2 = S2-S1 =2P+Q -P = P+Q∴ d = a2- a1 = P+Q-P = Q

Q27.

Answer :

Let there be two A.P.s.Let their first terms be a1 and a2 and their common differences be d1 and d2.Given: 5n+49n+6 = Sum of n terms in the first A.P.Sum of n terms in the second A.P.⇒5n+49n+6 = 2a1+[(n-1)d1]2a2+[(n-1)d2]Putting n=2×18-1= 35 in the above equation, we get: 5×35+49×35+6 = 2a1+34d12a2+34d2⇒179321= a1+17d1a1+17d1⇒179321= 18th term of the first A.P.18th term of the second A.P.

Q28.

Answer :

Given: ak = 5k+1For k=1, a1 = 5×1+1 = 6For k =2, a2 =5×2+1 = 11For k = n, an = 5n+1∴ Sn = n2a+an⇒Sn = n26+5n+1 = n25n+7

Q29.

Answer :

The two-digit numbers which when divided by 4 yield 1 as remainder are 13, 17….97.

∴ a =13, d = 4, an = 97∴ an = a+(n-1)d⇒97 = 13+(n-1)4⇒84 = 4n-4⇒88 = 4n⇒22= n …1Also, Sn = n22a+(n-1)d S22 = 2222×13+(22-1)×4 (From1)⇒S22 = 11110 = 1210

Q30.

Answer :

The given A.P. is 25,.22,.19…..
Here, a = 25, d = 22-25=-3
Sn =116⇒n22a+n-1d = 116⇒n2×25+n-1-3 = 232⇒50n-3n2+3n = 232⇒3n2 -53n+232 = 0

⇒3n2-29n-24n+232 = 0⇒n(3n-29)-8(3n-29) =0⇒(3n-29)(n-8) = 0⇒n = 293 or 8Since n cannot be a fraction, n = 8.Thus, the last term:an = a+(n-1)d⇒a8 =25+8-1×-3⇒a8 = 4

Q31.

Answer :

The odd integers from 1 to 2001 are 1, 3, 5…..2001.It is an AP with a=1 and d=2.an = 2001⇒1+(n-1)2=2001⇒2n-2=2000⇒2n=2002⇒n=1001Also, S1001 = 100122×1+1001-12⇒S1001 = 100122×1+10002⇒S1001 =10012×2002 = 1002001

Q32.

Answer :

Given:An A.P. with a =-6 and d = -112–6 = 12Sn = -25∴ -25 = n22×-6+n-112⇒-25 = n2-12+n2-12⇒-50 = nn2-252⇒-100 = nn-25⇒n2-25n+100 = 0⇒n-20n-5 = 0⇒ n =20 or n = 5

Q33.

Answer :

Given: a = 2, S5 = 14S10- S5We have: S5 =52 2×2+(5-1)d⇒S5 = 52+2d ….(i)Also, S10 = 1022×2+(10-1)d⇒S10 = 54+9d …..(ii)∵ S5 = 14S10- S5 From (i) and (ii), we have:⇒ 52+2d = 14 5(4+9d)-5(2+2d)⇒8+8d = 4+9d – 2-2d⇒d = -6∴ a20 = a+20-1d⇒a20 = a+19d⇒ a20 = 2+19-6⇒ a20 = -112

Page 19.36 Ex 19.5

Q1.

Answer :

Given: 1a,1b,1c are in A.P.∴ 2b = 1a+1c⇒2ac = ab+bc ….(1)(i) To prove: b+ca, c+ab,a+bc are in A.P. 2a+cb = b+ca+a+bc⇒2ac(a+c) = bc(b+c) +ab(a+b)LHS: 2ac(a+c)=(ab+bc)(a+c) (From(1))= a2b+2abc+bc2 RHS: bc(b+c) +ab(a+b)= b2c+bc2 +a2b+ab2= b2c+ab2+bc2 +a2b=b(bc+ab)+bc2+a2b= 2abc+bc2+a2b =a2b+2abc+bc2 (From(1))∴ LHS= RHSHence, proved.

(ii) To prove: a(b+c), b(c+a), c(a+b) are in A.P.⇒2b(c+a) =a(b+c)+c(a+b)LHS: 2b(c+a)= 2bc+2baRHS: a(b+c)+c(a+b)= ab+ac+ac+bc= ab+2ac+bc=ab+ab+bc+bc (From(1))= 2ab+2bc∴LHS = RHSHence, proved.

Q2.

Answer :

a2, b2, c2 are in A.P.∴ 2b2 = a2+c2⇒b2 -a2 = c2-b2⇒(b+a)(b-a) = (c-b)(c+b)⇒b-ac+b = c-bb+a⇒b-a(c+a)(c+b)= c-b(b+a)(c+a) Multiplying both the sides by 1c+a⇒1c+a-1b+c = 1a+b-1c+a∴’ 1b+c,1c+a,1a+b are in A.P.Multiplying each term by (a+b+c):a+b+cb+c, a+b+cc+a, a+b+ca+b are in A.P.Thus, ab+c+1 , bc+a+1 ,ca+b+1 are in A.P.Hence, ab+c, bc+a, ca+b are in A.P.

Q3.

Answer :

Since a, b, c are in A.P., we have:2b = a+c(i) We have to prove the following:2b2(a+c) = a2(b+c)+c2(a+b)LHS: 2b2 × 2b (Given) = 4b3RHS: a2b+a2c+ac2+c2b= ac(a+c) +b(a2+c2)= ac(a+c) +b[(a+c)2-2ac]=ac(2b)+b2b2-2ac=2abc +4b3-2abc=4b3LHS = RHS Hence, proved.

(ii) We have to prove the following:2(c+a-b)=(b+c-a)+(a+b-c)LHS: 2(c+a-b)=2(2b-b) ∵2b=a+c=2bRHS: (b+c-a)+(a+b-c)=2bLHS=RHSHence, proved.

(iii) We have to prove the following:2(ca-b2)=bc-a2+ab-c2RHS: bc-a2+ab-c2=c(b-c)+a(b-a)=ca+c2-c+aa+c2-a ∵ 2b=a+c=ca+c-2c2+aa+c-2a2=ca-c2+ac-a2=ca2-c22+ac2-a22=ac-12c2+a2=ac-124b2-2ac ∵ a2+c2+2ac=4b2⇒a2+c2=4b2 -2ac=ac-2b2+ac=2ac-2b2=2ac-b2=LHSHence, proved.

Q4.

Answer :

(i) Since b+ca,c+ab,a+bc are in A.P., we have:

c+ab – b+ca = a+bc – c+ab⇒ac+a2-b2-bcab = ab+b2-c2-acbc⇒a+ba-b+ca-bab = b+cb-c+ab-cbc⇒a-ba+b+cab = b-ca+b+cbc⇒a-bab = b-cbc⇒1b-1a = 1c-1b

Hence, 1a,1b,1c are in A.P.
(ii) Since 1a,1b,1c are in A.P., we have:

1b-1a=1c-1b⇒a-bab=b-cbc⇒a-ba=b-cc⇒a-bc = ab-c⇒ac – bc = ab – ac

Hence, bc, ca, ab are in A.P.

Q5.

Answer :

Since a, b, c are in A.P., we have:
2b = a+c
⇒b = a+c2

(i) Consider RHS:
4 (a − b) (b − c)
Substituting b=a+c2: ⇒4a-a+c2a+c2-c⇒42a-a-c2a+c-2c2⇒a-ca-c⇒a-c2
Hence, proved.
(ii) Consider RHS:
2 (ab + bc + ca)
Substituting b=a+c2:⇒2aa+c2+ca+c2+ac⇒2a2+ac+ac+c2+2ac2⇒a2+4ac+c2Hence, proved.
(iii) Consider RHS:
8b3
Substituting b=a+c2:⇒8a+c23⇒a3+c3+3aca+c⇒a3+c3+3ac(2b)⇒a3+c3+6abc

Hence, proved.

Q6.

Answer :

Given:
a1b+1c, b1c+1a, c1a+1b are in A.P.
By adding 1 to each term, we get: a1b+1c+1, b1c+1a+1, c1a+1b+1 are in A.P. ⇒a1b+1c+1a, b1c+1a+1b, c1a+1b+1c are in A.P.Dividing all terms by 1a+1b+1c, we get:⇒a, b, c are in A.P. Hence, proved.

Page 19.40 Ex 19.6

Q1.

Answer :

(i) Let A1 be the A.M. between 7 and 13.

A1 = a+b2

= 7+132

= 10

(ii) Let A1 be the A.M. between 12 and −8.

A1 = a+b2
= 12+-82
= 2

(iii) Let A1 be the A.M. between (x − y) and (x + y).
A1 = a+b2
= (x-y)+(x+y)2
= x

Q2.

Answer :

Let A1, A2, A3, A4 be the four A.M.s between 4 and 19. Then, 4, A1, A2, A3, A4 and 19 are in A.P. whose common difference is as follows:
d = 19-44+1
= 3

A1=4+d =4+3=7A2=4+2d=4+6=10A3=4+3d=4+9=13A4=4+4d=4+12=16
Hence, the required A.M.s are 7, 10, 13, 16.

Q3.

Answer :

Let A1, A2, A3, A4, A5, A6, A7 be the seven A.M.s between 2 and 17.
Then, 2, A1, A2, A3, A4, A5, A6, A7 and 17 are in A.P. whose common difference is as follows:

d = 17-27+1

=158

A1=2+d =2+158=318A2=2+2d =2+154=234A3=2+3d =2+458=618A4=2+4d =2+152=192A5=2+5d =2+758=918A6=2+6d =2+454=534A7=2+7d =2+1058=1218

Hence, the required A.M.s are 318, 234, 618, 192, 918, 534, 1218.

Q4.

Answer :

Let A1, A2, A3, A4, A5, A6 be the 6 A.M.s between 15 and -13.
Then, 15, A1, A2, A3, A4, A5, A6 and -13 are in A.P. whose common difference is as follows:

d = -13-156+1
= -4

A1=15+d =15+-4=11A2=15+2d =15+-8=7A3=15+3d =15+-12=3A4=15+4d =15+-16=-1A5=15+5d =15+-20=-5A6=15+6d =15+-24=-9
Hence, the required A.M.s are 11, 7, 3, -1, -5, -9.

Q5.

Answer :

Let A1, A2, A3, A4 ….An be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3, A1, A2, A3, A4….An and17.
Then, we have:
d = 17-3n+1 = 14n+1

Now, A1= 3 + d = 3 + 14n+1 = 3n+17n+1

And, An = 3+nd = 3+n14n+1 = 17n+3n+1

∴ AnA1=31⇒17n+3n+13n+17n+1=31⇒17n+33n+17=31⇒17n+3=9n+51⇒8n=48⇒n=6

Q6.

Answer :

Let A1, A2, A3, A4, 27, A6…An be the n arithmetic means between 7 and 71.
Thus, there are n+2 terms in all.
Let d be the common difference of the above A.P.
Now, a₆ = 27
⇒a+6-1d=27⇒a+5d=27
⇒d = 4

Also, 71 = an+2
71 = 7+n+2-14
⇒ 71 = 7+n+14
⇒ n = 15

Therefore, there are 15 A.M.s.

Q7.

Answer :

Let A1, A2……An be n A.M.s between two numbers a and b. Then, a, A1, A2…….An, b are in A.P. with common difference, d =b-an+1.∴ A1+A2 +……+An=n2A1+An =n2A1-d+An+d =n2a+b =n×a+b2 =A.M. between a and b, which is constant.

Q8.

Answer :

x, y, z are in A.P.
∴ y=x+z2

Now, A1 is the arithmetic mean of x and y.
A1=x+y2=x+x+z22=3x+z4

And, A2 is the arithmetic mean of y and z.
A2=y+z2=x+z2+z2=3z+x4

Let A3 be the arithmetic mean of A1 and A2.

A3=A1+A22 =3x+z4+3z+x42 =4x+4z8 =x+z2 =y

Hence, proved.

Q9.

Answer :

Let A1, A2, A3, A4, A5 be five numbers between 8 and 26.
Let d be the common difference.
Then, we have:
26 = a7
⇒26 = 8 + 7-1d
⇒26 = 8 + 6d
⇒d = 3

A1=8+d =8+3=11A2=8+2d=8+6=14A3=8+3d=8+9=17A4=8+4d=8+12=20A5=8+5d=8+15=23

Therefore, the five numbers are 11, 14, 17, 20, 23.

Page 19.42 Ex 19.7

Q1.

Answer :

Let the amount saved by the man in the first year be Rs A.
Let d be the common difference.
Let S10 denote the amount he saves in ten years.
Here, n =10, d =100

We know:
Sn=n22A+n-1d∴ S10=1022A+10-1100⇒16500 = 52A+900⇒3300 = 2A+900⇒A = 1200

Therefore, the man saved Rs 1200 in the first year.

Q2.

Answer :

Let n be the time in which the man saved Rs 200.
Here, d= 4, a = 32
We know:
Sn=n22a+n-1d⇒200 = n22×32+n-14⇒400 = 64n+4n2-4n⇒4n2+60n-400=0⇒n2+15n-100=0⇒n2+20n-5n-100=0⇒n+20n-5=0⇒n=5, n=-20 Rejecting the negative value

Therefore, the man took 5 years to save Rs 200.

Q3.

Answer :

Let S40 denote the total loan amount to be paid in 40 annual instalments.
∴ S40 = 3600
Let Rs a be the value of the first instalment and Rs d be the common difference.
We know:

Sn=n22a+n-1d

⇒4022a+40-1d = 3600⇒202a+39d = 3600⇒2a+39d = 180 ……1

Also, S40-S30=13×3600⇒3600 – S30=1200⇒S30=2400⇒3022a+30-1d=2400⇒152a+29d=2400⇒2a+29d=160 …..2

On solving equations 1 and 2, we get:
d = 2 and a =51

Hence, the value of the first instalment is Rs 51

Q4.

Answer :

Let an denote the production of radio sets in the nth year.
Here, a3 = 600, a7 = 700
We know:
an=a+n-1d

a3=a+2d⇒600 = a+2d …..1

And, a7=a+6d⇒700=a+6d …..2

Solving 1 and 2, we get:
d = 25, a = 550
Hence, the production in the first year is 550 units.

(ii) Let Sn denote the total production in n years.
Total production in 7 years = S7
=722×550 + 7-125=4375 units

(iii) Production in the 10th year = a10
a10=a+10-1d =550+925 =775

Q5.

Answer :

Let Sn be the total distance travelled by the gardener.
Let d be the common difference (distance) between two trees. Let a be the distance of the well from the first tree.
Here, n = 25, d = 10, a = 20
Distance travelled by the gardener from the well to the last tree = S25
S25=2522×20 + 25-110 =25240+240 =3500 m

Therefore, the total distance the gardener has to travel is 3500 m.

Page 19.43 Ex 19.7

Q6.

Answer :

It is given that the man counts Rs 180 per minute for half an hour.
∴ Sum of money the man counts in 30 minutes = Rs 180×30 = Rs 5400

Total money counted by the man = Rs 10710
∴ Money left for counting after 30 minutes = Rs (10710 − 5400) = Rs 5310

It is given that after 30 minutes, he counts at the rate of Rs 3 less every minute than the preceding minute.

Therefore, it would be an A.P. where a = 177 and d = −3.
Let the time taken to count Rs 5310 be n minutes.
5310=n22×177+n-1×-3⇒10620=354 n-3n2+3n⇒3n2-357n+10620=0⇒n2-119n+3540=0⇒n2-59n-60n+3540=0⇒nn-59-60n-59=0⇒n-59n-60=0∴ n=59 or 60

Thus, the time taken to count Rs 5310 would be 59 minutes or 60 minutes.

Hence, the total time taken to count Rs 10710 would be (30 + 59) minutes or (30 + 60) minutes, i.e. 89 minutes or 90 minutes, respectively.

Q7.

Answer :

Cost of the piece of equipment at the end of the first year = Rs 600000 – 15% of 600000
= Rs 600000 – Rs 90000
= Rs 510000
Cost of the piece of equipment at the end of the second year = Rs 600000 – 13.5% of 600000
= Rs 600000 – Rs 81000
= Rs 519000
Cost of the piece of equipment at the end of the third year = Rs 600000 – 12% of Rs 600000
= Rs 600000 – Rs 72000
= Rs 528000
The numbers 510000, 519000, 528000 are in an A.P. whose first term in 510000 and common difference is 9000.
∴ Cost of the piece of equipment at the end of 10 years = a + (10 – 1)d
= 510000 + 81000
= Rs 591000

Q8.

Answer :

Cost of the tractor = Rs 12000
It is given that the farmer pays Rs 6000 in cash.
Unpaid amount = Rs 6000
He has to pay Rs 6000 in annual instalments of Rs 500 plus 12% interest on the unpaid amount.
∴ Number of years taken by the farmer to pay the whole amount = 6000 ÷ 500 = 12
Hence, the interest paid by farmer annually would be as follows:
12% of Rs 6000+12% of Rs 5500+12% of Rs 5000 ..=720+660+600….

It is in an A.P. where a = 720 , d = -60 and n = 12.
Total sum:
1222×720+11×-60=61440-660=Rs 4680

∴ Amount the farmer has to pay = Rs 12000 + Rs 4680 = Rs 16680

Q9.

Answer :

Cost of the scooter = Rs 22000
Shamshad Ali pays Rs 4000 in cash.
∴ Unpaid amount = Rs 22000 – Rs 4000 = Rs 18000
Number of years taken by Shamshed Ali to pay the whole amount = 18000 ÷ 1000 = 18
He agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount.

Total amount of instalments:
10% of Rs 18000+10% of Rs 17000+10% of Rs 16000….=1800+1700+1600….

It is in an A.P. where a = 1800, d = -100 and n = 18.
Therefore, total amount of instalments:
1822×1800+(18-1)×-100=93600-1700=Rs 17100

∴ Total amount Shamshad Ali has to pay = Rs (22000 + 17100) = Rs 39100

Q10.

Answer :

Let Sn denote the total amount the person receives in n years.
Let d be the common increment in his income every year.
Let a denote the initial income of the person.
Here, a = 300,000, d = 10000, n = 20

Total amount at the end of 20 years:
S20=2022×300,000+(20-1)10,000 =79,00,000

Therefore, the total amount the person receives in 20 years is Rs 79,00,000.

Q11.

Answer :

Let a30 be the amount a man repays in the 30th instalment.
Let d be the common increment in his instalment every month.
Let a be the initial repayment.
Here, a = 100, d = 5, n = 30
Amount to be repaid in the 30th instalment:
a30⇒a+n-1d
=100+29×5=245

Hence, the man repays Rs 245 in his 30th instalment.

Page 19.43 (Very Short Answers)

Q1.

Answer :

We have: an=xn+y∴ a1=x+ya2=2x+y
Common difference of an A.P., d = a2-a1
⇒2x+y – x+y⇒x

Q2.

Answer :

Sum of the first n terms of an A.P. = p2n2+Qn
Sum of one term of an A.P. = S1
⇒p212+Q1⇒p2+Q

Sum of two terms of an A.P. = S2
⇒p222+Q2⇒2p+2Q
Now, we have:
a1+a2=S2⇒p2+Q+ a2 = 2p+2Q⇒a2 = Q+32p

Common difference:
d=a2-a1 =Q+32p-Q+p2 =p

Q3.

Answer :

Given:
Sn=2n2+3n
⇒S1=212+31 =5S2 = 222+32 =14∴ a1+a2=14 ⇒5+a2 =14 ⇒a2 = 9

Common difference, d = a2-a1
= 9 – 5
= 4
nth term = a +n-1d
= 5+n-14
= 4n+1

Q4.

Answer :

The numbers log 2, log (2^x − 1) and log (2^x + 3) are in A.P.

∴log 2x-1-log 2 = log 2x+3-log 2x-1⇒log 2x-12=log 2x+32x-1⇒2x-12=2x+32x-1⇒2x-12=22x+3⇒22x+1-2.2x=2.2x+6⇒22x-4.2x-5=0

Let 2x=y.⇒y2-4y-5=0⇒y-5y+1=0⇒y=5 or y=-1∴2x=5 or 2x=-1 not possibleTaking log on both the sides, we get: log 2x=log5⇒xlog2=log5⇒x=log 5 log 2=log2 5⇒x=log2 5

Q5.

Answer :

Given:

SnSn1=2n+53n+4⇒n22a+n-1dn22b+n-1d1=2n+53n+4⇒2a+n-1d2b+n-1d1=2n+53n+4 …1

Ratio of their m terms = ambm
To find the ratio of the mth terms, replace n by 2m-1 in equation (1).
⇒2a+2m-2d2b+2m-2d1=22m-1+532m-1+4⇒a+m-1db+m-1d1=4m-2+36m-3+4⇒ambm=4m+16m+1

Q6.

Answer :

We need to find the sum of 1, 3, 5, 7… upto n terms.
Here, a = 1, d = 2

We know:
Sn=n22a+n-1d =n22×1+n-12 =n2

Therefore, the sum of the first n odd numbers is n2.

Q7.

Answer :

We need to find the sum of 2, 4, 6, 8…upto n terms.
Here, a = 2, d = 2

We know:
Sn=n22a+n-1d =n22×2+n-12 =nn+1

Therefore, the sum of the first n odd numbers is nn+1.

Q8.

Answer :

For the first series, a = 3, d1= 7
For the second series, b = 63, d2= 2

Given:
an=bn⇒a+n-1d1=b+n-1d2⇒3+n-17=63+n-12⇒3+7n-7=63+2n-2⇒5n=65⇒n=13

Hence, the 13th terms of both the series are the same.

Q9.

Answer :

3+5+7+…+upto n terms5+8+11+…. upto 10 terms=7⇒n22×3+n-121022×5+10-13=7⇒n4+2n370=7⇒n2+2n-1295=0⇒n2+37n-35n-1295=0⇒n+37n-35⇒n=35, n=-37Rejecting the negative value, we get:⇒n=35

Q10.

Answer :

Given:

am=n⇒a+m-1d=n ….1an=m⇒a+(n-1)d=m ….2

Solving equations 1 and 2, we get:
d = -1
a = n+m-1

pth term:
ap=a+p-1d = n+m-1 +p-1-1 =n+m-p

Hence, the pth term is n + m -p.

Page 19.44 (Multiple Choice Questions)

Q1.

Answer :

a7=34⇒a+6d=34 …..1Also, a13=64⇒a+12d=64 …..2

Solving equations 1 and 2, we get:

a = 4 and d = 5

∴a18 = a+17d =4+175 =89

Q2.

Answer :

(d) -p+q

Sp=q⇒p22a+p-1d=q⇒2ap+p-1pd=2q …..1Sq=p⇒q22a+q-1d=p⇒2aq+q-1qd=2p …..2Multiplying equation 1 by q and equation 2 by p and then solving, we get:d=-2p+qpqNow, Sp+q=p+q22a+p+q-1d =p22a+p-1d+qd + q22a+q-1d+pd =Sp+pqd2+Sq+pqd2 =p+q+pqd =p+q-2p+qpqpq =-(p+q)

Q3.

Answer :

(a) 2

Sn=3n2-n⇒S1=312-1⇒S1=2∴ a1= 2

Q4.

Answer :

(b) 1210

The given series is 13, 17, 21….97.

a1=13, a2=17, an=97d=a2-a1=7-3=4

an=97⇒a+n-1d=97⇒13+n-14=97⇒n=22

Sum of the above series:
S22= 2222×13+22-14 =1126+84 =1210

Q5.

Answer :

(a) 6
Let A1, A2, A3, A4 …. An be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3, A1, A2, A3, A4, …. An and 17.
Then, we have:

d = 17-3n+1 = 14n+1

Now, A1= 3 + d = 3 +14n+1 = 3n+17n+1

And, An = 3+nd = 3+n14n+1 = 17n+3n+1

∴ AnA1=31⇒17n+3n+13n+17n+1=31⇒17n+33n+17=31⇒17n+3=9n+51⇒8n=48⇒n=6

Q6.

Answer :

(b) 2 m-12 n-1

SmSn=m2n2⇒m22a+m-1dn22a+n-1d=m2n2⇒2a+m-1d2a+n-1d=mn ⇒2an+ndm-nd=2am+nmd-md⇒2an-2am-nd+md=0⇒2an-m-dn-m=0⇒2an-m=dn-m⇒d=2a …..1
Ratio of aman=a+m-1da+n-1d⇒ aman=a+m-2aa+n-2a From (1) = a+2am-2aa+2an-2a =2am-a2an-a = 2m-12n-1

Q7.

Answer :

(b) 6

a= 1, an=11, Sn= 36∵ an= 11⇒a+n-1d= 11⇒1+n-1d= 11⇒n-1d= 10 …..1Also, Sn= 36⇒n22a+n-1d=36⇒n2+10=72 Using (1)⇒n=6

Q8.

Answer :

(b) 27th

Sn=3n2+5nS1=312+51 = 8∴ a1=8S2=322+52 = 22∴ a1+a2=22⇒ a2 = 14Common difference, d =14-8= 6Also, an=164⇒a+n-1d=164⇒8+n-16=164⇒n=27

Q9.

Answer :

Sn=2n2+5nS1=2.12+5.1 = 7∴ a1=7Sn=2.22+5.2=18∴ a1+a2=18⇒ a2=11Common difference, d =11-7= 4an=a+n-1d =7+n-14 =4n+3

Q10.

Answer :

(c) cot a₁ − cot a_n
We have:
sin d cosec a1 cosec a2+cosec a2 cosec a3+….+cosec an-1 cosec an =sin dsin a1 sin a2+sin dsin a2 sin a3+…..+sin dsin an-1 sin an=sin (a2-a1)sin a1 sin a2+ sin (a3-a2)sin a2 sin a3+….+sin (an-an-1)sin an-1 sin an=sin a2 cos a1-cos a2 sin a1sin a1 sin a2+sin a3 cos a2-cos a3 sin a2sin a1 sin a2+…..+sin a2 cos a1-cos a2 sin a1sin a1 sin a2=cot a1-cot a2 +cot a2-cot a3+…..+cot an-1-cot an=cot a1-cot an

Q11.

Answer :

(a) 1/5

S3n=S4n-S3n⇒2S3n=S4n⇒2×3n22a+3n-1d = 4n22a+4n-1d⇒32a+3n-1d = 22a+4n-1d⇒6a+9nd-3d=4a+8nd-2d⇒2a+nd-d=0⇒2a+n-1d=0 ….1

Required ratio: S2nS4n-S2n

S2nS4n-S2n = 2n22a+2n-1d4n22a+4n-1d-2n22a+2n-1d =nnd2n3nd-nnd =16-1 =15

Q12.

Answer :

(d) tan a_n − tan a_{1Answer :
We have:}

sin d sec a1 sec a2+sec a2 sec a3+….+sec an-1sec an =sin dcos a1cos a2+sin dcos a2 cos a3+…..+sin dcos an-1 cos an=sin (a2-a1)cos a1cos a2+ sin (a3-a2)cos a2 cos a3+….+sin (an-an-1)cos an-1 cos an=sin a2 cos a1-cos a2 sin a1cosa1 cos a2+sin a3 cos a2-cos a3 sin a2cos a1 cos a2+…..+sin a2 cos a1-cos a2 sin a1cos a1 cos a2=tan a1-tan a2 +tan a2-tan a3+…..+tan an-1-tan an=tan a1-tan ana3+…..+sin dcos an-1 cos an=sin (a2-a1)cos a1cos a2+ sin (a3-a2)cos a2 cos a3+….+sin (an-an-1)cos an-1 cos an=sin a2 cos a1-cos a2 sin a1cosa1 cos a2+sin a3 cos a2-cos a3 sin a2cos a1 cos a2+…..+sin a2 cos a1-cos a2 sin a1cos a1 cos a2=tan a1-tan a2 +tan a2-tan a3+…..+tan an-1-tan an=tan a1-tan an
Q13.

Answer :

(a) 5, 10, 15, 20

Let the four numbers in A.P. be as follows:
a-2d, a-d, a, a+d
Their sum = 50 (Given)
⇒a-2d+a-d +a+a+d = 50⇒2a-d=25 …..1Also, a+d = 4a-2d⇒a+d=4a-8d⇒3d=a ……2

From equations 1 and 2, we get:
d = 5, a = 15

Hence, the numbers are 15-10, 15-5, 15, 15+5, i.e. 5, 10, 15, 20.

Page 19.45 (Multiple Choice Questions)

Q14.

Answer :

(d) 14

The given series is 1, . . . . . . . . . . . , 31
There are n A.M.s between 1 and 31: 1, A1, A2, A3, . . . . ., An, 31.

Common difference, d = 31-1n+1 = 30n+1

Here, we have:
A1An=329⇒1+d1+nd=329⇒1+30n+11+n×30n+1=329⇒n+1+30n+1+30n=329⇒n+3131n+1=329⇒29n+899=93n+3⇒64n=896⇒n=14

Q15.

Answer :

(b) 2

Let the A.P. be a, a+d, a+2d, a+3d…
Given:
d=Sn-kSn-1+Sn-2
For n = 3, we have:

d=3a+3d-k2a+d+a⇒4a+2d-k2a+d=0⇒22a+d=k2a+d⇒2=k

Q16.

Answer :

(b) 2S

Given:

S=n2l+a⇒l+a=2Sn

Also, d=l2-a2k-l+a⇒d=l+al-ak-l+a⇒d=n-1d×2Snk-2Sn⇒k-2Sn=n-12Sn⇒k=2Snn-1+1⇒k=2S

Q17.

Answer :

(d) n+1n

Given:
Sum of the even natural numbers = k × Sum of the odd natural numbers
n22a+n-1d=k×n22a+n-1d⇒2×2+n-12=k×2×1+n-12⇒4+n-122+n-12=k⇒n+1n=k

Q18.

Answer :

Let the A.P. be a, a+d, a+2d……..a+nd.
Here, let d be the common difference and n be the total number of terms.

Given:
a1=a,a2=b⇒a+d=b⇒d=b-a …..1an=2a⇒a+n-1d=2a⇒n-1d=a⇒d=an-1 …..2

From equations 1 and 2, we have:
⇒an-1=b-a⇒ab-a+1=n⇒a+b-ab-a=n⇒bb-a=n

Now, sum of n terms of an A.P.:
S = n2a+an =n23a =3ab2b-a

Q19.

Answer :

(a) 2nn+1
Let n be an odd number.
Given:

S1=Sum of odd number of terms =n22a+n-1d …..1Since n is odd, the number of odd places = n+12S2=Sum of the terms of a series in odd places =n+1222a+n+12-12d =n+142a+n-1d …..2

From equations 1 and 2, we have:

S1S2=n22a+n-1dn+142a+n-1d =2nn+1

Q20.

Answer :

Given:
Sn=n2p⇒n22a+n-1d=n2p⇒2a+n-1d=2np⇒2a=2np-n-1d …..1Sm=m2p⇒m22a+m-1d=m2p⇒2a+m-1d=2mp⇒2a=2mp-m-1d …..2

From 1 and 2, we have:

2np-n-1d=2mp-m-1d⇒2pn-m=dn-1-m+1⇒2p=d

Substituting d = 2p in equation 1, we get:
a = p

Sum of p terms of the A.P. is given by:

p22a+p-1d=p22p+p-12p =p3

GEOMETRIC PROGRESSION

Page 20.10 Ex 20.1

Q1.

Answer :

i We have,a1= 4, a2 =-2, a3 =1, a4 =-12Now, a2a1=-24=-12, a3a2=1-2, a4a3=-121=-12∴ a2a1=a3a2=a4a3=-12Thus, a1, a2, a3 and a4 are in G.P., where a=4 and r=-12.

ii We have,a1=-23 , a2 =-6, a3 =-54Now, a2a1=-6-23=9, a3a2=-54-6=9 ∴ a2a1=a3a2=9Thus, a1, a2 and a3 are in G.P., where a=-23 and r=9.

iii We have,a1=a , a2 =3a24, a3 =9a316Now, a2a1=3a24a=3a4, a3a2=9a3163a24=3a4 ∴ a2a1=a3a2=3a4Thus, a1, a2 and a3 are in G.P., where the first term is a and the common ratio is 3a4.

iv We have,a1=12 , a2 =13, a3 =29, a4 =427Now, a2a1=1312=23, a3a2=2913=23, a4a3= 42729=23∴ a2a1=a3a2=a4a3=23Thus, a1, a2, a3 and a4 are in G.P., where the first term is 12 and the common ratio is 23.

Q2.

Answer :

We have,an=23n, n∈NPutting n=1, 2, 3, …a1=231=23, a2=232=29, a3=233=227 and so on.Now, a2a1=2923=13, a3a2=22729=13 and so on.∴a2a1=a3a2= … =13So, the sequence is an G.P., where 23 is the first term and 13 is the common ratio.

Q3.

Answer :

i) Here, First term, a=1 Common ratio, r=a2a1=41=4∴9th term=a9=ar(9-1)=1(4)8=48=65536Thus, the 9th term of the given GP is 65536.

ii) Here, First term, a=-34 Common ratio, r =a2a1=12-34=-23∴10th term=a10=ar(10-1)=-34-239=12238Thus, the 10th term of the given GP is 12238.

iii) Here, First term, a=0.3Common ratio, r =a2a1=0.060.3=0.2∴8th term=a8=ar(8-1)=0.3(0.2)7Thus, the 8th term of the given GP is 0.3(0.2)7.

iv) Here, First term, a=1a3x3Common ratio, r =a2a1=ax1a3x3=a4x4∴12th term=a12=ar(12-1)=1a3x3(a4x4)11=a41x41Thus, the 12th term of the given GP is a41x41.
v) Here, First term, a=3Common ratio, r =a2a1=133=13∴nth term=an=ar(n-1)=313n-1Thus, the nth term of the given GP is 313n-1.

viHere, First term, a=2Common ratio, r =a2a1=122=12∴10th term=a10=ar(10-1)=2129=12×128Thus, the 10th term of the given GP is 12×128.

Q4.

Answer :

Here, first term, a=227Common ratio, r=a2a1=29227=3 Last term, l=162After reversing the given G.P., we get another G.P. whose first term is l and common ratio is 1r.∴ 4th term from the end = l1r4-1=(162)133=6

Q5.

Answer :

We have,a2a1=0.020.004=5, a3a2=0.10.02=5⇒a2a1=a3a2=5The given progression is a G.P. whose first term, a is 0.004 and common ratio, r is 5.Let the nth term be 12.5.∴ an=12.5⇒ arn-1=12.5⇒(0.004)(5)n-1=12.5⇒(5)n-1=12.50.004⇒(5)n-1=3125⇒(5)n-1=(5)5Comparing the power of both the sides⇒n-1=5⇒n=6Thus, 6th term of the given G.P. is 12.5

Q6.

Answer :

i Here, first term, a=2 and common ratio, r=12Let the nth term be 15122. ∴ an = 15122⇒ arn-1 = 15122⇒ 212n-1 = 15122⇒ 12n-1 = 11024⇒ 12n-1 = 1 210 ⇒n-1 = 10 ⇒n=11Thus, the 11th term of the given G.P. is 15122.

ii Here, first term, a=2 and common ratio, r=2Let the nth term be 128. ∴ an = 128⇒ arn-1 = 128⇒ 22n-1 = 128⇒2 (2)n-1 = 128⇒ 2n-1 = 64⇒ 2n-1 = 212 ⇒n-1 = 12 ⇒n = 13Thus, the 13th term of the given G.P. is 128.

iii Here, first term, a=3 and common ratio, r=3Let the nth term be 729. ∴ an=729⇒ arn-1 = 729⇒ 33n-1 = 729⇒ (3)n-1 = 3123=(3)11⇒n-1 = 11⇒n=12Thus, the 12th term of the given G.P. is 729.

(iv) Here, first term, a=13 and common ratio, r=13Let the nth term be 119683. ∴ an=119683⇒ arn-1 = 119683⇒ 1313n-1 = 119683⇒ 13n-1 = 339=138⇒ n-1 = 8 ⇒n=9Thus, the 9th term of the given G.P. is 119683.

Q7.

Answer :

Here, first term, a = 18 and common ratio, r = -23Let the nth term be 512729.∴ arn-1 = 512729⇒18-23n-1=512729⇒-23n-1=512729×118 =2566561⇒-23n-1=-238⇒n-1 = 8 ⇒n=9Thus, the 9th term of the given G.P. is 512729.

Q8.

Answer :

After reversing the given G.P., we get another G.P. whose first term, l is 14374 and common ratio is 3.∴ 4th term from the end=l1r4-1 =14374 34-1=274374= 1162

Q9.

Answer :

Let a be the first term and r be the common ratio of the given G.P. ∴ a4 =27 and a7 = 729⇒ar3 = 27 and ar6 = 729⇒ar6ar3 = 72927⇒ r3 = 33 ⇒ r = 3Putting r = 3 in ar3 = 27a33 = 27 ⇒a = 1Thus, the given G.P. is 1, 3, 9, …

Q10.

Answer :

Let a be the first term and r be the common ratio.∴ a7=8a4 and a5=48⇒ar6=8ar3 and ar4=48⇒r3= 8 ⇒r3=23⇒r = 2Putting r = 2 in ar4=48a24 = 48 ⇒a=3Thus, the given G.P. is 3, 6, 12, …

Q11.

Answer :

Given:First term, a=5 Common ratio, r=2an= 52n-1 … 1Similarly, an= 128012n-1 … 2From 1 and 252n-1 = 128012n-1⇒1256=14n-1 ⇒144=14n-1 ⇒ n-1 = 4 ⇒n = 5

Q12.

Answer :

a2+b2+c2p2-2ab+bc+cdp+b2+c2+d2≤0⇒a2p2+b2p2+c2p2-2abp+bcp+cdp+b2+c2+d2≤0⇒a2p2-2abp+b2+b2p2-2bcp+c2+c2p2-2cdp+d2≤0⇒ap-b2+bp-c2+cp-d2≤0⇒ap-b2+bp-c2+cp-d2=0⇒ap-b2=0 ⇒p = baAlso, bp-c2=0 ⇒p = cbSimiliarly, ⇒cp-d2=0 ⇒p = dc∴ ba= cb=dcThus, a, b, c and d are in G.P.

Q13.

Answer :

Given:a+bxa-bx=b+cxb-cx=c+dxc-dxNow, a+bxa-bx=b+cxb-cxApplying componendo and dividendo⇒a+bx+a-bxa+bx-a-bx=b+cx+b-cxb+cx-b-cx⇒2a2bx = 2b2cx⇒ab=bcSimiliarly, b+cx+b-cxb+cx-b-cx=c+dx+c-dxc+dx-c-dx⇒ bc = cdTherefore, a, b, c and d are in G.P.

Q14.

Answer :

Let a be the first term and r be the common ratio of the given G.P.∴ p = 5th term ⇒p = ar4 … 1q = 8th term ⇒q = ar7 … 2s = 11th ⇒s = ar10 … 3Now, q2 = ar72 = a2r14⇒ar4 ar10 = ps From 1 and 3∴ q2 = ps

Q15.

Answer :

Let r be the common ratio of the given G.P. Then, a4 = a22 GivenNow, ar3 = a2r2 ⇒ r = a ⇒ r = -3 Putting a=-3 ∴ a7 = ar6 ⇒ a7 = -3 -36 Putting a=-3 and r=-3 ⇒ a7=-3-729 ⇒ a7=-2187Thus, the 7th term of the G.P. is -2187.

Q16.

Answer :

Let a be the first term and r be the common ratio.∴ a3 = 24 and a6 = 192⇒ar2 = 24 and ar5 = 192⇒ar5ar2=19224⇒r3 = 8 ⇒r3 =23 ⇒r=2Putting r=2 in ar2=24a22 = 24 ⇒a = 6Now, 10th term =a10 =ar9Putting a=6 and r=2 in a10 = ar9⇒a10= 629 =3072Thus, the 10th term of the G.P. is 3072.

Page 20.15 Ex 20.2

Q1.

Answer :

Let the terms of the the given G.P. be ar, a and ar.
Then, product of the G.P. = 3375
⇒ a³ = 3375
⇒a = 15
Similarly, sum of the G.P. = 65
⇒ar+a+ar = 65
Substituting the value of a
15r+15+15r=65⇒15r2+15r+15=65r⇒15r2-50r+15=0⇒53r2-10r+3=0⇒3r2-10r+3 = 0⇒3r-1r-3=0⇒r=13, 3
Hence, the G.P. for a = 15 and r = 13 is 45, 15, 5.
And, the G.P. for a = 15 and r = 3 is 5, 15, 45.

Q2.

Answer :

Let the terms of the the given G.P. be ar, a and ar.
Then, product of the G.P. = 1728
⇒ a³ = 1728
⇒a = 12
Similarly, sum of the G.P. = 38
⇒ar+a+ar = 38
Substituting the value of a
12r+12+12r=38⇒12r2+12r+12=38r⇒12r2-26r+12=0⇒26r2-13r+6=0⇒6r2-13r+6 = 0⇒3r-22r-3=0⇒r=23,32
Hence, the G.P. for a = 12 and r = 23 is 18, 12 and 8.
And, the G.P. for a = 12 and r = 32 is 8, 12 and 18.

Hence, the three numbers are 8, 12 and 18.

Q3.

Answer :

Let the first three numbers of the given G.P. be ar, a and ar.
∴ Product of the G.P. = −1
⇒ a³ = −1
⇒a = −1
Similarly, Sum of the G.P. = 1312
⇒ar+a+ar = 1312
Substituting the value of a = −1
-1r-1-r=1312⇒12r2+25r+12=0⇒12r2+16r+9r+12=0⇒4r3r+4+33r+4=0⇒4r+33r+4 = 0⇒r=-34,-43
Hence, the G.P. for a = −1 and r = -34 is 43,-1 and 34.

And, the G.P. for a = −1 and r = -43 is 34,-1 and 43.

Q4.

Answer :

Let the required numbers be ar, a and ar.
Product of the G.P. = 125
⇒a3=125 ⇒ a = 5
Sum of the products in pairs = 8712=1752

⇒ar×a+a×ar+ar×ar=1752⇒a2r+a2r+a2=1752Substituting the value of a⇒25r +25r+25=1752⇒50r2+50r+50 =175r⇒50r2-125r+50=0⇒25(2r2-5r+2)=0⇒2r2-4r-r+2=0⇒2r(r-2)-1(r-2)=0⇒(2r-1)(r-2)=0∴ r = 12, 2

Hence, the G.P. for a = 5 and r = 12 is 10, 5 and 52.
And, the G.P. for a = 5 and r = 2 is 52, 5 and 10.

Q5.

Answer :

Let the required numbers be a, ar and ar2.
Sum of the numbers = 21
⇒a+ar+ar2=21⇒a(1+r+r2)=21 …(i)
Sum of the squares of the numbers = 189
⇒a2+(ar)2+(ar2)2 = 189
⇒a2+(ar)2+(ar2)2 = 189 ⇒a21+r2+r4 = 189 …(ii)
Now, a ( 1 + r + r2) = 21 [From (i)]Squaring both the sides⇒a21 + r + r22= 441⇒a2 1 + r2+r4 + 2a2r1+r+r2 = 441⇒189 + 2ara1+r+r2 = 441 [Using (ii)]⇒189 +2ar×21 = 441 [Using (i)]⇒ar = 6 ⇒a=6r …(iii)Putting a = 6r in (i) 6r1+r+r2=21⇒6r+6+6r=21⇒6r2 +6r+6=21r⇒6r2-15r+6=0⇒3(2r2-5r+2)=0⇒2r2-5r+2 = 0⇒(2r-1)(r-2)=0⇒r = 12, 2Putting r = 12 in a=6r, we get a=12. So, the numbers are 12, 6 and 3.Putting r = 2 in a=6r, we get a=3. So, the numbers are 3, 6 and 12.Hence, the numbers that are in G.P are 3, 6 and 12.

Q6.

Answer :

Let the numbers be a, ar and ar².
Sum=14 ⇒a+ar+ar2=14 ⇒a(1+r+r2)=14 … i
According to the question, a + 1, ar + 1 and ar2 − 1 are in A.P.
∴ 2ar+1 = a+1 + ar2-1⇒2ar + 2 = a + ar2⇒2ar + 2 = 14-ar [Fromi]⇒3ar = 12 ⇒a = 4r … iiPutting a = 4r in i⇒4r(1+r+r2)=14⇒4r2-10r+4 = 0 ⇒4r2-8r-2r+4 = 0 ⇒4r-2r-2=0⇒r=12, 2
Putting r = 12in ii, we get a = 8.So, the G.P. is 8, 4 and 2.

Similarly putting r = 2 in (ii), we get a = 2.
So, the G.P is 2, 4 and 8.

Q7.

Answer :

Let the terms of the given G.P. be ar, a and ar.
∴ Product = 216
⇒a3=216⇒a=6
It is given that ar+2, a+8 and ar+6 are in A.P.
∴ 2a+8=ar+2+ar+6Putting a = 6, we get⇒28 = 6r+2+6r+6⇒28r = 6r2+8r+6⇒6r2 -20r+6=0⇒6r-2r-3 =0⇒r = 13, 3Hence, putting the values of a and r, the required numbers are 18, 6, 2 or 2, 6 and 18.

Q8.

Answer :

Let the required numbers be ar, a and ar.
Product of the G.P. = 729
⇒a3=729⇒a=9
Sum of the products in pairs = 819
⇒ar×a+a×ar+ar×ar=819⇒a21r+r+1 = 819⇒811+r2+rr= 819⇒9r2+r+1=91r⇒9r2-82r+9=0⇒9r2-81r-r+9=0⇒9r-1r-9=0⇒r=19, 9Hence, putting the values of a and r, we get the numbers to be 81, 9 and 1 or 1, 9 and 81.

Q9.

Answer :

Let the terms of the G.P be ar, a and ar.
∴ Product of the G.P. = 1
⇒a3=1⇒a=1
Now, sum of the G.P. = 3910
⇒ar+a+ar=3910⇒a1r+1+r=3910⇒11r+1+r=3910⇒10r2+10r+10=39r⇒10r2-29r+10=0⇒10r2-25r-4r+10=0⇒5r(2r-5)-2(2r-5)=0⇒5r-22r-5=0⇒r=25, 52
Hence, putting the values of a and r , the required numbers are 52, 1, 25 or 25, 1 and 52.

Page 20.24 Ex 20.3

Q1.

Answer :

(i) Here, a = 2 and r = 3.
∴ S7 = ar7-1r-1 =2 37-13-1 =2187-1=2186

(ii) Here, a = 1 and r = 3.
∴ S8 = ar8-1r-1 =1 38-13-1 =6561-12=3280

(iii) Here, a = 1 and r = −12.
∴ S9 = a1-r91-r = 1 1–1291–12 =1–151232=51351232=513×2512×3=171256

(iv) Here, a = a² − b² and r = 1a+b.
∴ Sn = a1-rn1-r = a2-b2 1-1a+bn1-1a+b =a2-b2a+bn-1a+bna+b-1a+b⇒Sn=a+ba-ba+bn-1a+bn-1a+b-1=a-ba+bn-2a+bn-1a+b-1

(v) Here, a = 4 and r = 12
∴ Sn=a1-r101-r=41-12101-12=41-1102412=81-11024=1023128

Page 20.25 Ex 20.3

Q2.

Answer :

(i) Here, a = 0.15 and r =a2a1=0.0150.15=110.
S8=a1-r81-r =0.151-11081-110=0.151-1108110=161-1108

(ii) Here, a = 2 and r = 12.
S8=a1-r81-r=21-1281-12=21-125612=22255256=2552128

(iii) Here, a =29 and r = -32.
S5=ar5-1r-1=29-325-1-32-1=29-24332-1-32-1=29-27532-52= 11001440=5572

Q3.

Answer :

(i)
S11 = ∑n=1112+3n⇒S11 = ∑n=1112 + ∑n=1113n ⇒S11 = 2 × 11 + 3+32+33+ … +311= 22 +3311-13-1 =22 +177147-12= 22+265719 = 265741

(ii)
Sn =∑k=1n2k+3k-1=∑k=1n2k + ∑k=1n3k-1=2 + 4 + 8+ … +2n + 1 + 3 + 9+ … +3n =22n-12-1 + 13n-13-1 = 122n+2-4+3n-1 = 122n+2+3n-5

(iii)
∑n=2104n= 42 + 43+44+ … +410=16 + 64 +256+ … + 410=1649-14-1 = 16349-1
Q4.

Answer :

(i) We have,
5 + 55 + 555+ … n terms
Taking 5 as common:
Sn = 5[1 + 11 + 111 + … n terms]
= 599+99+999+ … n terms= 5910-1+102-1+103-1+ … +10n-1= 5910+102+103+ … +10n – 1+1+1+1+ … n times= 5910 × 10n-110-1 – n = 59 10910n-1 – n= 58110n+1-9n-10

(ii) We have,
7 + 77 + 777 + … n terms
Sn= 7 [1 + 11 + 111 + … n terms]
= 799+99+999+ … n terms= 7910-1+102-1+103-1+ … +10n-1= 7910+102+103+ … +10n – 1+1+1+1 … n times= 7910 × 10n-110-1 – n = 79 10910n-1 – n= 78110n+1-9n-10

(iii) We have,
9 + 99 + 999 + … n terms
= 9+99+999+ … + to n terms= 10-1+102-1+103-1+ … +10n-1= 10+102+103+ … +10n – 1+1+1+1 … n times= 10 × 10n-110-1 – n = 10910n-1 – n= 1910n+1-9n-10

(iv) We have,
0.5 + 0.55 + 0.555 + … n terms
Sn= 5 [0.1 + 0.11+0.111 + … n terms]
= 590.9+0.99+0.999+ … + to n terms= 59910+9100+91000+ … n terms= 591-110+1-1100+1-11000+ … n terms = 59n-110+1102+1103+ … n terms = 59n-1101-110n1-110= 59n-191-110n

(v) We have,
0.6 + 0.66 +.666 + … to n terms
Sn= 6 [0.1 + 0.11+ 0.111 + … n terms]
= 690.9+0.99+0.999+ … n terms= 69910+9100+91000+ … n terms= 691-110+1-1100+1-11000+ … n terms = 69n-110+1102+1103+ … n terms = 69n-1101-110n1-110= 69n-191-110n

Q5.

Answer :

Here, a = 3
Common ratio, r = 12
Sn = 3069512
∴ S_n= 31-12n1-12⇒3069512 = 31-12n12 ⇒3069512= 6 1-12n⇒30693072=1-12n ⇒ 12n= 1 – 30693072 ⇒ 12n= 33072⇒2n = 30723⇒2n=1024 ⇒2n= 210∴ n = 10

Q6.

Answer :

Here, a = 2
Common ratio, r = 3
Sum of n terms, S_n = 728
Sn=23n-13-1 ⇒728 = 23n-12⇒728 =3n-1 ⇒3n = 729⇒3n= 36∴ n = 6

Q7.

Answer :

Here, a = 3
Common ratio, r = 3
Sum of n terms, S_n = 39+33

Sn=33n-13-1 ⇒39+133 = 33-13n-1⇒3n-1 = 39+1333-13⇒3n = 1 + 26⇒3n = 27 ⇒3n = 36∴ n = 6

Q8.

Answer :

Here, a = 3
Common ratio, r = 3
Sum of n terms, S_n = 381
∴ Sn = 3 + 6 + 12 + … + n terms
⇒381 = 32n-12-1 ⇒381 = 3 2n-1⇒127 = 2n-1⇒2n=128 ⇒ 2n = 27 ∴ n = 7

Q9.

Answer :

Here, common ratio, r = 3
nth term, a_n = 486
S_n = 728

an=486 ⇒arn-1=486⇒a3n-1 = 486 ⇒a3n = 486 × 3 ⇒a3n =1458 … iNow, Sn = 728⇒728 = a 3n-13-1 ⇒728 = a3n – a2⇒1456 = a3n-1 – a ⇒1456= 1458- a From i⇒a = 1458 – 1456 ⇒a= 2

Q10.

Answer :

Let a be the first term and r be the common ratio of the G.P.

∴ S3 =ar3-1r-1 and S6 = ar6-1r-1Then, according to the question S3S6=ar3-1r-1 a r6-1r-1 ⇒125152 = r3-1r6-1⇒125 r6-1 = 152 r3-1⇒125r6-125 = 152r3 – 152⇒125r6 – 152r 3+ 27 = 0Now, let r3 = y ∴ 125y2 – 152y + 27 = 0Now, applying the quadatic formulay = -b±b2-4ac2a ⇒ y = 152±9604250⇒ y = 152+9604250 or 152-9604250⇒y = 1 or 27125∴ r3 = 1 or r3 = 27125But, r = 1 is not possible.∴ r = 271253 = 35

Q11.

Answer :

Let a be the first term and r be the common ratio of the G.P.

∴ a4=127 ⇒ar4-1 = 127⇒ar3 = 127 ⇒ ar32 = 1272⇒a2r6 = 1729 ⇒ ar6 = 1729a … iSimilarly, a7=1729 ⇒ ar7-1 = 1729⇒ar6 = 1729 ⇒ar6= 1729a From i ∴ a = 1Putting this in a4=127⇒ar3 = 133⇒r3 = 133 ∴ r = 13Now, sum of n terms of the G.P., Sn = arn-1r-1⇒Sn = 11-13n1-13 ⇒Sn= 321-13n

Q12.

Answer :

∑n=11012n-1 + 15n+1=∑n=11012n-1 + ∑n=11015n+1 = 1+12+14+ … +129 + 152+153+154+ …+1511=11-12101-12 + 1251-15101-15 =210-129 + 510-14×511

Q13.

Answer :

Let a be the first term and r be the common ratio of the G.P.

a2=24 ⇒ar2-1=24⇒ar = 24 …iSimilarly, a5 = 81 ⇒ar5-1 = 24⇒ar4 = 81⇒24×r4r= 81 From i⇒r3 = 8124 ∴ r3 =278⇒r =32Putting r =32 in i3a = 48 ⇒a =16So, the geometric series is 16+ 24 + 36+ …+16328And, S8=16328-132-1 ⇒S8=326561-256256=32×6305256=63058

Page 20.26 Ex 20.3

Q14.

Answer :

Let a be the first term and r be the common ratio of the given G.P.

Sum of n terms, S1=arn-1r-1 … 1Sum of 2n terms, S2 =ar2n-1r-1⇒S2 =arn2-12r-1⇒S2 =arn-1rn+1r-1⇒S2 =S1rn+1 ….2And, sum of 3n terms, S3= ar3n-1r-1⇒S3=arn3-13r-1⇒S3=arn-1r2n+rn+1r-1⇒S3=S1r2n+rn+1 …3

Now, LHS=S12 + S22 =S12 + S1rn+12 Using 2=S121 + rn+12=S121+r2n+2rn+1=S12r2n+rn+1+rn+1=S1S1r2n+rn+1+S1rn+1=S1S2+S3 Using 2 and 3=RHSHence proved.

Q15.

Answer :

Given: S1, S2, …, Sn are the sum of n terms of an G.P. whose first term is 1 in each case and the common ratios are 1, 2, 3, …, n.∴ S1=1+1+1+ … n terms = n … 1 S2=12n-12-1 =2n-1 … 2 S3=13n-13-1 =3n-12 … 3 S4=14n-14-1 =4n-13 … 4…. Sn=1nn-1n-1 =nn-1n-1 …………nNow, LHS= S1+S2+2S3+3S4 + … +n-1Sn=n+2n-1+3n-1+4n-1+ … +nn-1 Using 1, 2, 3, …, n=n+2n+3n+4n+ … +nn-1+1+1+ … +n-1 times=n+2n+3n+4n+ … +nn-n-1=n+2n+3n+4n+ … +nn-n+1=1+2n+3n+4n+ … +nn=1n+2n+3n+4n+ … +nn=RHSHence proved.

Q16.

Answer :

Let there be 2n terms in the given G.P. with the first term being a and the common ratio being r.
According to the question
Sum of all the terms = 5 (Sum of the terms occupying the odd places)
⇒a1 + a2 + … +a2n = 5 a1+a3+a5+ … +a2n-1⇒a + ar + … +ar2n-1 = 5 a+ar2 + … +ar2n-2⇒a1-r2n1-r = 5a1-r2n1-r2 ⇒ 1+r = 5 ∴ r = 4

Q17.

Answer :

Let a be the first term and r be the common ratio of the G.P.

∴ ∑n=1100a2n = α and ∑n=1 100a2n-1 =β∴ a2+a4+ … +a200 = α and a1+a3+ … +a199 = β⇒ar + ar3 + … +ar199 =α and a+ar2+ … +ar198 = β⇒ar1-r21001-r2 = α and a1-r21001-r2 = βNow, dividing α by βαβ = ar1-r21001-r2 a1-r21001-r2 = arr=r∴ r = αβ

Q18.

Answer :

Let the given series be a1 + a2 + a3 + a4+ … +a2n.Now, it is given that a1 = 1, a2 = aa1, a3 = ca2, a4 = aa3, a5 =ca4 and so on.∵ a1 = 1⇒a1 = 1, a2 = a, a3 = ac, a4 = a2c, a5=a2c2, a6 = a3c2,…..∴ Sum of the 2n terms of the series, Sn= a1 + a2 + a3 + a4 + … + a2n= 1 + a + ac +a2c + a2c2 + … +2n terms=1+a + ac1+a +a2c21+a + … + n terms=1+a1-acn1-ac = 1+a acn-1ac-1

Q19.

Answer :

Let a be the first term and r be the common ratio of the G.P.
Sum of the first n terms of the series = a1+a2+a3+ … +an
Similarly, sum of the terms from n+1th to 2nth term = an+1 +an+2 + … + a2n
∴ Required ratio = a1+a2+a3+ … +an an+1 +an+2 + … +a2n = a +ar+ … +arn-1arn + arn+1+ … +ar2n-1=a1-rn1-rarn1-rn1-r = 1rn

Q20.

Answer :

We have,
a +b = 3, ab = p, c + d =12 and cd = q
a, b, c and d form a G.P.
∴ First term = a, b = ar, c = ar² and d = ar³
Then, we have
a + b = 3 and c + d = 12
⇒ a+ar = 3 ⇒ a( 1+ r ) = 3 …iSimilarly, ar2(1+r) = 12 …ii⇒ar21+ra1+r = 123⇒r2 = 4 ⇒r = 2∴ a 1+r = 3 ⇒ a = 1Now, p = ab ⇒ p = a × ar = 2And, q = cd ⇒q = ar2 × ar3 = 25 = 32∴q+pq-p=32+232-2=3430=1715

Q21.

Answer :

Here, a = 3 and Common ratio, r =12 And, Sn = 3069512∴Sn = 31-12n1-12⇒3069512=31-12n1-12 ⇒3069512= 6 1-12n⇒30693072 = 1 -12n ⇒12n = 1 – 30693072 ⇒12n=33072⇒2n = 30723 ⇒2n= 1024 ⇒2n= 210∴ n = 10

Q22.

Answer :

Here, the ancestors of the person form the G.P. 2, 4, 8, 16, ……..
Now, first term, a = 2
And, r = 2
∴ Number of his ancestors during the ten generations preceding his own, S10 = 2210-12-1 = 2 1024-1=2046

Page 20.35 Ex 20.4

Q1.

Answer :

i In the given G.P., first term, a=1 and common ratio, r=-13Hence, the sum S to infinity is given by S = a1-r=11–13=34.ii In the given G.P., first term, a=8 and common ratio, r=12Hence, the sum S to infinity is given by S = a1-r=81-12=2+2.iii We have: 25+352+253+354 . = 25+253+ ..∞+352+354+ …=An infinite G.P. with a= 25 and r=125+An infinite G.P.with a=325 and r=125= 251-125+ 351-125= 252425+352425=1024+324=1324

Q2.

Answer :

LHS=913.919.9127 … ∞=913+19127.=9131-13=9131-13=9=3=RHS

Q3.

Answer :

LHS=214.428.8316.16432 …∞=214+28+316316432.∞=2122+223+324+425 + …∞=21221+22+322+423 …∞=212211-12+1.121-122=21222+2=21=2=RHS

Q4.

Answer :

We have:Sp= 1+rp+r2p+ … ∞∴ Sp=11-rpSimilarly, sp= 1-rp+r2p- … ∞∴ sp=11–rp=11+rpNow, SP+sp = 11-rp+11+rp = 1-rp+1+rp1-r2p⇒21-r2p = 2S2P∴ SP+sp = 2S2P

Q5.

Answer :

Let r be the common ratio of the given G.P.∴ a = 4Sum of the geometric ifinite series:S∞= 4+4r+4r2+ … ∞Now, S∞ = 41-r …….iThe difference between the third and fifth term is 3281.a3 -a5 = 3281⇒4r2-4r4 = 3281⇒ 4r2-r4=3281⇒81r4-81r2+8 =0 …….iiNow, let r2 = yLet us put this in ii.∴ 81r4-81r2+8 =0⇒81y2-81y+8 = 0⇒81y2-72y-9y+8 = 0⇒9y9y-1-89y-1 = 0⇒9y-89y-1⇒y=19, 89Putting y = r2, we get r=13 and 223Substituting r=13 and 223 in i:S∞= 41-13=122=6Similarly, S∞ = 41-223=123-22∴ S∞ = 6, 123-22

Q6.

Answer :

Here, first term, a = 1
Common ratio = r

∴ an = an+1 + an+2+ an+3+…..∞ ∀ n∈N⇒arn-1 = arn+ arn-1+…..∞⇒rn-1 = rn1-r Putting a =1⇒rn-1 1-r = rn ⇒1-r = r⇒2r = 1 ⇒ r =12Thus, the infinte G.P is 1, 12, 14, … ∞.

Q7.

Answer :

Let the first term be a and the common difference be r.

∴ a1 + a2 = 5 ⇒ a + ar = 5 …iAlso, an= 3 an+1+an+2+an+3+ … ∞ ∀ n∈ N⇒arn-1 = 3 arn+1+arn+2+arn+3+ … ∞⇒arn-1 = 3arn1-r ⇒1-r = 3r⇒4r = 1 ⇒r =14Putting r =14 in i:a + a4 = 5⇒5a = 20 ⇒ a = 4Thus, the G.P. is 4, 1, 14, 116, …∞.

Q8.

Answer :

Let the rational number S be 0.125.∵ S=0.125=0.125+0.000125+0.000000125+0.000000000125+… ∞⇒S=0.1251+10-3+10-6+10-9 + … ∞Clearly, S is a geometric series with the first term, a, being 1 and the common ratio, r, being 10-3.∴ S=11-r⇒S=0.12511-10-3⇒S=125999

Q9.

Answer :

Let the rational number S be 0.423.∵ S=0.423=0.4+0.023+0.00023+0.0000023+ … ∞⇒S=0.4+0.0231+10-2+10-4+ … ∞Clearly, S is a geometric series with the first term, a, being 1 and the common ratio, r, being 10-2.∴ S=0.4+0.02311-10-2⇒S=0.4+2.399⇒S=419990

Q10.

Answer :

Let us take a G.P. with terms a1, a2, a3, a4, … ∞and common ratio rr<1.Also, let us take the sum of all the terms following each term to be S1, S2, S3, S4, …Now, S1=a21-r=ar1-r,S2=a31-r=ar21-r,S3=a41-r=ar31-r,…⇒a1S1=aar1-r=1-rr,a2S2=arar21-r=1-rr,a3S3=ar2ar31-r=1-rr,…It is clearly seen that the ratio of each term to the sum of all the terms following it is constant.

Q11.

Answer :

(i) 0.3Let S=0.3⇒S=0.3+0.03+0.003+0.0003+0.00003+ … ∞⇒S=0.31+10-1+10-2+10-3+10-4+ … ∞S is a geometric series with the first term, a, being 1 and the common ratio, r, being 10-1.∴ S=11-r⇒S=0.311-10-1⇒S=39=13

(ii) 0.231Let S = 0.231⇒S=0.231+0.000231+0.000000231+ … ∞⇒S=0.2311+10-3+10-6+ … ∞It is a G.P.∴ S=0.23111-10-3⇒S=231999

(iii) 3.52Let S=3.52⇒S=3.5+0.02+0.002+0002+0.00002+ … ∞⇒S=3.5+0.021+10-1+10-2+10-3+10-4+ … ∞It is a G.P. ∴ S=3.5+0.0211-10-1⇒S=3.5+0.29⇒S=31790

(iv) 0.68Let S=0.68⇒S=0.6+0.08+0.008+0.0008+0.00008+ … ∞⇒S=0.6+0.081+10-1+10-2+10-3+ … ∞It is a G.P.∴ S=0.6+0.0811-10-1⇒S=0.6+0.89⇒S=6.29⇒S=6290=3145

Page 20.36 Ex 20.4

Q12.

Answer :

According to the midpoint theorem, the sides of each triangle formed by joining the midpoints of an equilateral triangle are half of the sides of the equilateral triangle. In other words, the triangles formed are equilateral triangles with sides 18 cm, 9 cm, 4.5 cm, 2.25 cm, …

(i) Sum of the perimeters of all the triangles, P= 3×18+3×9+3×4.5+3×2.25+… ∞⇒P=3×18+9+4.5+2.25+… ∞It is a G.P. with a=18 and r=12.∴ P=3×181-12⇒P=3×36=108 cm

(ii) Sum of the areas of all the triangles, A=34182+3492+344.52+ … ∞⇒A=34182+92+4.52+ … ∞It is a G.P. with a= 182 and r=14.∴ A=34 1821-14⇒A=33×324⇒A=1083 cm2

Q13.

Answer :

S=a1-r …….(i)And, S1=a21-r2 ⇒ S1=a21-r1+r …….(ii)Now, putting the value of a in equation (ii) from equation (i):S1=S21-r21-r1+r⇒S1=S21-r1+r⇒S11+r=S21-r⇒rS1+S2=S2-S1⇒r=S2-S1S1+S2Putting the value of r in equation (i):⇒a=S1-r⇒a=S1-S2-S1S1+S2⇒a=SS1+S2-S2-S1S1+S2⇒a=2SS1S1+S2

Page 20.41 Ex 20.5

Q1.

Answer :

a ,b and c are in G.P.
∴ b2 = ac
Now, taking log on both the sides:⇒logb2 = log ac ⇒2log b = log a + log cThus, log a, log b and log c are in A.P.

Q2.

Answer :

a, b, c are in G.P.
∴ b2 = ac Now taking logm on both the sides:⇒logmb2 = logmac⇒2logmb = logma + logmc⇒2logbm = 1logam +1logcmThus, 1logam, 1logbm and 1logcm are in A.P

Q3.

Answer :

a, b and c are in A.P.∴ 2b=a+c …….(i)Also, a, b and d are in G.P.∴ b2=ad …….(ii)Now, a-b2=a2-2ab+b2=a2-aa+c+ad Using (i) and (ii)=ad-ac=ad-c⇒a-b2=ad-cTherefore, a, a-b and d-c are in G.P.

Q4.

Answer :

Here, ap=a + p-1daq= a + q-1dar = a + r-1d as= a + s-1dIt is given that ap, aq, ar and as are in G.P.∴ aqap=araq=aq-arap-aq=q-rp-q …….(i)Similarly, araq=asar=ar-asaq-ar=r-sq-r …….(ii)Using i and ii:q-rp-q=r-sq-r, Therefore, p-q, q-r and r-s are in G.P.

Q5.

Answer :

Here, 1a+b, 12b and 1b+c are in A.P.

∴ 2×12b=1a+b+1b+c⇒1b=b+c+a+ba+bb+c⇒a+bb+c =b2b+a+c⇒ab+ac+b2+bc = 2b2+ab+bc⇒2b2 – b2 = ac⇒b2 = ac Thus, a, b and c are in G.P.

Q6.

Answer :

Here, xa= xzb2 = zcNow, taking log on both the sides:logxa=logxzb2=logzc⇒alog x =b2 logxz = c log z⇒alog x =b2log x + b2log z = c log z⇒alog x =b2log x + b2log z andb2log x +b2log z= c log z⇒a-b2log x=b2 log z andb2log x=c-b2log z⇒log x log z=b2a-b2 and log x log z=c-b2b2 ⇒b2a-b2=c-b2b2⇒b24=ac-ab2-bc2+b24⇒2ac=ab+bc⇒2b = 1a+1cThus, 1a, 1b and 1c are in A.P.

Q7.

Answer :

k, k + 9, k−6 are in G.P.
∴ k-62=4k+9⇒k2+36-12k = 4k + 36⇒k2 – 16k = 0⇒k k-16 = 0⇒k=0, 16
But, k = 0 is not possible.
∴ k = 16

Q8.

Answer :

Let the first term of an A.P. be a and its common difference be d.

a1+a2+a3=15⇒a+a+d+a+2d=15⇒3a + 3d = 15 ⇒ a+d = 5 …….(i)Now, according to the question:a + 1, a+d+3 and a+2d+9 are in G.P.⇒a+d+32 = a+1a+2d+9⇒5-d+d+32 = 5-d+1 5-d+2d+9 From (i) ⇒82 = 6-d14+d⇒64 = 84 + 6d-14d -d2⇒d2+8d-20=0⇒d-2d+10 = 0⇒d=2, -10Now, putting d = 2, -10 in equation (i), we get, a = 3, 15, respectively.Thus, for a = 3 and d=2, the A.P. is 3, 5, 7.And, for a = 15 and d=-10, the A.P. is 15 , 5, -5.

Q9.

Answer :

Let the first term of an A.P is a and its common difference be d.

∴ a1+a2+a3=21⇒a+a+d+a+2d=21⇒3a + 3d = 21 ⇒ a+d = 7 …(i)Now, according to the question:a , a+d-1 and a+2d+1 are in G.P.⇒a+d-12 = aa+2d+1⇒7+a-a-12 = a a+27-a+1 ⇒62 = a15-a⇒36 = 15a -a2⇒a2-15a+36=0⇒a-3a-12 = 0⇒a=3, 12Now, putting a = 2, 12 in equation (i), we get d = 5, -5, respectively.Thus, for a = 2 and d=5, the numbers are 2, 7 and 12.And, for a = 12 and d=-5, the numbers are 12 ,7 and 2.

Q10.

Answer :

Let the first term of the A.P. be a and the common difference be d.
∴ a = a , b = a + d and c = a + 2d
a+b+c=18⇒a+a+d+a+2d=18⇒3a + 3d = 18 ⇒ a+d = 6 …….(i)Now, according to the question, a + 4, a+d+4 and a+2d+36 are in G.P.∴ a+d+42 = a+4a+2d+36⇒6-d+d+42 =6-d+4 6-d+2d+36 ⇒102 = 10-d42+d⇒100 = 420 + 10d-42d -d2⇒d2+32d-320=0⇒d+40d-8 = 0⇒d=8, -40Now, putting d = 8, -40 in equation (i), we get, a = -2, 46, respectively.For a = -2 and d=8, we have: a = -2 , b = 6 , c = 14And, for a = 46 and d=-40, we have: a = 46 , b = 6 , c = -34

Q11.

Answer :

Let the first term of a G.P be a and its common ratio be r.
∴ a1+a2+a3=56⇒a+ar+ar2 = 56 ⇒a 1+r+r2 = 56 ⇒a=561+r+r2 …….i Now, according to the question:a-1, ar-7 and ar2-21 are in A.P.∴ 2ar-7 = a-1 + ar2 – 21⇒2ar – 14 = ar2+a-22⇒ar2-2ar+a-8 = 0⇒a1-r2 = 8⇒a = 81-r2 …….iiEquating (i) and (ii):⇒81-r2=561+r+r2⇒81+r+r2 = 561+r2-2r ⇒1+r+r2 =7 1+r2-2r⇒1+r+r2 =7+7r2-14r⇒6r2-15r+6=0 ⇒ 32r2-5r+2 = 0⇒2r2-4r-r+2=0⇒2r(r-2)-1(r-2)=0⇒(r-2)(2r-1)=0⇒r=2, 12When r=2, a=8. [Using (ii)]And, the required numbers are 8, 16 and 32.When r=12, a=32. [Using (ii)]And, the required numbers are 32, 16 and 8.

Q12.

Answer :

a, b and c are in G.P.
∴ b2=ac …….(1)

(i) LHS=ab2+c2=ab2+ac2=aac+cb2 Using (1)=ca2+b2=RHS

(ii) LHS=a2b2c21a3+1b3+1c3=b2c2a+a2c2b+a2b2c=acc2a+b22b+a2acc Using (1)=a3+b3+c3=RHS

(iii) LHS=a+b+c2a2+b2+c2=a+b+c2a2-b2+c2+2b2=a+b+c2a2-b2+c2+2ac Using (1)=a+b+c2a+b+ca-b+c ∵ a+b+ca-b+c=a2-b2+c2+2ac=a+b+ca-b+c =RHS

(iv) LHS=1a2-b2+1b2=b2+a2-b2a2-b2b2=a2a2b2-b4=a2a2ac-ac2=1ac-c2=1b2-c2=RHS

(v) LHS =a+2b+2ca-2b+2c=a2-4b2+4c2+4ac=a2-4ac+4c2+4ac Using (1)=a2+4c2=RHS

Q13.

Answer :

a, b, c and d are in G.P.

∴ b2=acbc=adc2=bd …….(1)

(i) LHS=ab-cdb2-c2=ab-cdac-bd Using (1)=ab-cdbac-bdb =ab2-bcdac-bdb=aac-cc2ac-bdb Using (1)=a2c-c3ac-bdb=ca2-c2ac-bdb=a+cac-c2ac-bdb=a+cac-bdac-bdb Using (1)=a+cb=RHS

(ii) LHS=a+b+c+d2=a+b2+2a+bc+d+c+d2=a+b2+2ac+ad+bc+bd+c+d2=a+b2+2b2+bc+bc+c2+c+d2 Using (1)=a+b2+2b+c2+c+d2=RHS

(iii) LHS=b+cb+d=b2+bd+bc+cd=ac+c2+ad+cd Using (1)=ca+c+da+c=c+ac+d = RHS

Q14.

Answer :

a, b and c are in G.P.
∴ b2=ac …….(1)
(i) b22=ac2 Using (1)⇒b22=a2c2Therefore, a2, b2 and c2 are also in G.P.

(ii) b32=b23=ac3 Using (1)⇒b32=a3c3Therefore, a3, b3 and c3 are also in G.P.

(iii) ab+bc2=ab2+2ab2c+bc2⇒ab+bc2=ab2+ab2c+ab2c+bc2⇒ab+bc2=a2b2+acac+b2b2+b2c2 Using (1)⇒ab+bc2=a2b2+c2+b2b2+c2⇒ab+bc2=b2+c2a2+b2Therefore, a2+b2, b2+c2 and ab+bc are also in G.P.

Page 20.42 Ex 20.5

Q15.

Answer :

a, b, c and d are in G.P.

∴ b2=acad=bc c2=bd …….(1)

(i) b2+c22=b22+2b2c2+c22⇒b2+c22=ac2+b2c2+b2c2+bd2 Using (1)⇒b2+c22=a2c2+a2d2+b2c2+b2d2 Using (1)⇒b2+c22=a2c2+d2+b2c2+d2 ⇒b2+c22=a2+b2c2+d2Therefore, a2+b2, c2+d2 and b2+c2 are also in G.P.

(ii) b2-c22=b22-2b2c2+c22⇒b2-c22=ac2-b2c2-b2c2+bd2 Using (1)⇒b2-c22=a2c2-b2c2-a2d2+b2d2 Using (1)⇒b2-c22=c2a2-b2-d2a2-b2⇒b2-c22=a2-b2c2-d2Therefore, a2-b2, b2-c2 and c2-d2 are also in G.P.

(iii) 1b2+c22=1b22+2b2c2+1c22⇒1b2+c22=1ac2+1b2c2+1b2c2+1bd2 Using (1)⇒1b2+c22=1a2c2+1a2d2+1b2c2+1b2d2 Using (1)⇒1b2+c22=1a21c2+1d2+1b21c2+1d2⇒1b2+c22=1a2+b21c2+1d2Therefore, 1b2+c2, 1c2+d2 and 1b2+c2 are also in G.P.

(iv) ab+bc+cd2=ab2+bc2+cd2+2ab2c+2bc2d+2abcd⇒ab+bc+cd2=a2b2+b2c2+c2d2+ab2c+ab2c+bc2d+bc2d+abcd+abcd⇒ab+bc+cd2=a2b2+b2c2+c2d2+b2b2+acac+c2c2+bdbd+bcbc+adad Using (1)⇒ab+bc+cd2=a2b2+a2c2+a2d2+b4+b2c2+b2d2+c2b2+c4+c2d2⇒ab+bc+cd2=a2b2+c2+d2+b2b2+c2+d2+c2b2+c2+d2⇒ab+bc+cd2=b2+c2+d2a2+b2+c2Therefore, a2+b2+c2, ab+bc+cd and b2+c2+d2 are also in G.P.

Q16.

Answer :

a-b, b-c and c-a are in G.P.∴ b-c2=a-bc-a⇒b2-2bc+c2=ac-bc+ab-a2⇒a2+b2+c2=ab+bc+ca …….(i)Now, LHS=a+b+c2=a2+b2+c2+2ab+2bc+2ca=ab+bc+ca+2ab+2bc+2ca Using (i)=3ab+3bc+3ca=3ab+bc+ca=RHS

Q17.

Answer :

a, b and c are in A.P.∴ 2b=a+c …….(i)Also, b, c and d are in G.P.∴ c2=bd …….(ii)And 1c, 1d and 1e are in A.P.∴ 2d=1c+1e ⇒d=2cec+e …….(iii)∵ c2=bd From (ii) ⇒c2=a+c 22cec+e Using (i) and (iii)⇒c2c+e=cea+c⇒c2+ce=ae+ec⇒c2=aeTherefore, a, c and e are also in G.P.

Q18.

Answer :

a, b and c are in A.P.∴ 2b=a+c …….(i)a, x and b are in G.P.∴ x2=ab …….(ii)And, b, y and c are also in G.P.∴ y2=bc …….(iii)Now, putting the values of a and c:⇒2b=x2b+y2b⇒2b2=x2+y2Therefore, x2, b2 and y2 are also in A.P.

Q19.

Answer :

a, b and c are in A.P.∴ 2b=a+c …….(i)Also, a, b and d are in G.P.∴ b2=ad …….(ii)Now, a-b2=a2-2ab+b2⇒ a-b2=a2-aa+c+ad Using (i) and (ii)⇒ a-b2=a2-a2-ac+ad⇒ a-b2=ad-ac⇒ a-b2=a(d-c)Therefore, a, a-b and (d-c) are in G.P.

Q20.

Answer :

a, b and c are in G.P.∴ b2=ac ……..(i)Now, LHS=a2+ab+b2bc+ca+ab=a2+ab+acbc+b2+ab Using (i)=aa+b+cbc+b+a=ab=1rHere, r = common ratioRHS=b+ac+b=ar+aar2+ar=a(r+1)ar(r+1)=1r∴ LHS = RHS

Q21.

Answer :

Let r be the common ratio of the given G.P.∴ b=ar and c=ar2Now, a+b+c=bx⇒a+ar+ar2=arx⇒r2+1-xr+1=0 r is always a real number.∴ D≥0⇒1-x2-4≥0⇒x2-2x-3≥0⇒x-3x+1≥0⇒x>3 or x<-1 and x≠3 or -1 ∵ a, b and c are distinct real numbers

Q22.

Answer :

a4=x⇒ar3=xAlso, a10=y⇒ar9=yAnd, a16=z⇒ar15=z∵ yx=ar9ar3=r6and zy=ar15ar9=r6∴yx=zyTherefore, x, y and z are in G.P.

Page 20.48 (Very Short Answers)

Q1.

Answer :

Here, a₅ = 2
⇒ar4 = 2
Product of the nine terms, i.e. a, ar, ar2, ar3, ar4, ar5, ar6, ar7 and ar8:
a×ar8ar×ar7ar2×ar6ar3×ar5ar4 = ar49∵ar4 = 2Required product = 29 = 512

Q2.

Answer :

Here, p + qth term = m ⇒ arp + q-1 = m …….iAnd, p – qth term = n ⇒ arp – q-1 = n …….iiDividing i by ii:arp + q-1arp – q-1=mn ⇒r2q = mn⇒rq = mnNow, from i:arp-1×rq = m⇒arp-1 × mn = m⇒arp-1 = m ×nm⇒arp-1 = mnmThus, the pth term is mnm.

Q3.

Answer :

logx a, ax2 and logb x are in G.P.∴ ax22 = logx a × logb x ⇒ax = logba logbx× logbx ⇒ax = logba Now, by taking loga on both the sides:⇒ x = logalogba

Page 20.48 Ex 20.6

Q1.

Answer :

Let the 6 G.M.s between 27 and 181 be G1, G2, G3, G4, G5 and G6.Thus, 27, G1, G2, G3, G4, G5, G6 and 181 are in G.P.∴ a=27, n=8 and a8=181∵ a8=181⇒ar7=181⇒r7=181×27⇒r7=137⇒r=13 ∴ G1=a2=ar=2713=9G2=a3=ar2=27132=3G3=a4=ar3=27133=1G4=a5=ar4=27134=13G5=a6=ar5=27135=19 G6=a7=ar6=27136=127

Q2.

Answer :

Let the 5 G.M.s betweem 16 and 14 be G1, G2, G3, G4 and G5.16, G1, G2, G3, G4, G5, 14⇒a=16, n=7 and a7=14∵ a7=14⇒ar6=14⇒r6=14×16⇒r6=126⇒r=12∴ G1=a2=ar=1612=8G2=a3=ar2=16122=4G3=a4=ar3=16123=2G4=a5=ar4=16124=1G5=a6=ar5=16125=12

Q3.

Answer :

Let the 5 G.M.s between 329 and 812 be G1, G2, G3, G4 and G5.329, G1, G2, G3, G4, G5, 812⇒a=329, n=7 and a7=812∵ a7=812⇒ar6=812⇒r6=812×932⇒r6=326⇒r=32∴ G1=a2=ar=32932=163G2=a3=ar2=329322=8G3=a4=ar3=329323=12G4=a5=ar4=329324=18G5=a6=ar5=329325=27

Q4.

Answer :

(i) Let the G.M. between 2 and 8 be G.Then, 2, G and 8 are in G.P.∴ G2=2×8⇒G2=16⇒G=±16⇒G=±4

(ii) Let the G.M. between a3b and ab3 be G.Then, a3b, G and ab3 are in G.P.∴ G2=a3b×ab3⇒G2=a4b4⇒G=a4b4⇒G=a2b2

(iii) Let the G.M. between -8 and -2 be G.Then, -8, G and -2 are in G.P.∴ G2=-8-2⇒G2=16⇒G=±16⇒G=4, -4

Q5.

Answer :

a is the G.M. between 2 and 14.∴ a2=2×14⇒a2=12⇒a=12⇒a=12

Q6.

Answer :

Let G1, G2, G3, G4, …, Gn be n G.M.s between a and b.Then, a, G1, G2, G3, G4, …, Gn, b is a G.P.Let r be the common ratio.∵ b=an+2=arn+1⇒r=ba1n+1∴ G1=a2=arG2=a3=ar2G3=a4=ar3…Gn=an+1=arnAlso, let G be the G.M. between a and b.∴ G2=abNow, G1×G2×G3×G4× … × Gn = ar×ar2×ar3×ar4× … ×arn=an×r1+2+3+4+……+n=an×rnn+12=an×ba1n+1nn+12=an×ban2=an2×bn2=abn2=abn=Gn∴ G1×G2×G3×G4× … × Gn=Gn

Q7.

Answer :

Let A.M. and G.M. between the two numbers a and b be A and G, respectively. A=25⇒a+b2=25⇒a+b=50 …….(i)Also G= 20⇒ab=20⇒ab=400 …….(ii)Now, putting the value of a in (ii):⇒(50-b)b=400⇒b2-50b+400=0⇒b2-10b-40b+400=0⇒bb-10-40b-10=0⇒b-10b-40=0⇒b=10, 40If b=10, then, a=400.And, if b=40, then a=10.Thus, the two numbers are 10 and 40.

Q8.

Answer :

AM=2GM∴ a+b2=2ab⇒a+b=4abSquaring both the sides:⇒a+b2=4ab2⇒a2+2ab+b2=16ab⇒a2-14ab+b2=0Using the quadratic formula:⇒a=–14b+-14b2-4×1×b22×1 ∵ a is positive number⇒a=14b+2b49-12⇒a=b7+43⇒ab=7+43⇒ab=4+3+2×2×3⇒ab=2+32⇒ab=2+322-32-3⇒ab=2+34-32-3∴ ab=2+32-3

Q9.

Answer :

Let the two positive numbers be a and b.a, A and b are in A.P.∴ 2A=a+b ……..(i)Also, a, G1, G2 and b are in G.P.∴ r=ba13Also, G1=ar and G2=ar2. ……..(ii)Now, LHS=G12G2+G22G1=ar2ar2+ar22ar Using (ii)=a+ar3=a+aba133=a+aba=a+b=2A=RHS Using (i)

Q10.

Answer :

Let the roots of the quadratic equation be a and b. A=a+b2∴ a+b=2A ……..(i)Also, G2=ab …….(ii)The quadratic equation having roots a and b is given by x2-(a+b)x+ab=0.∴ x2-2Ax+G2=0 Using (i) and (ii)

Q11.

Answer :

Let the two numbers be a and b.Let the geometric mean between them be G.We have:a+b=6GBut, G=ab∴ a+b=6ab⇒a+b2=6ab2⇒a2+2ab+b2=36ab⇒a2-34ab+b2=0Using the quadratic formula:⇒a=–34b±-34b2-4×1×b22×1⇒a=34b±b1156-42⇒a=b34±11522⇒ab=34±2422⇒ab=17+122 ∵ a and b are positive numbers⇒ab=3+8+2×3×22⇒ab=3+222⇒ab=3+2223-223-22⇒ab=3+229-83-22⇒ab=3+223-22⇒a:b=3+22:3-22

Q12.

Answer :

Let the roots of the quadratic equation be a and b. AM=8∴ a+b2=8⇒a+b=16 ……..(i)Also, G=5⇒ab=5⇒ab=52⇒ab=25 ……..(ii)Now, the quadratic equation is given by x2-(a+b)x+ab=0⇒x2-16x+25=0

Q13.

Answer :

AM=10∴ a+b2=10⇒a+b=20 ……..(i)Also, G=8∴ ab=8⇒ab=82⇒ab=64 ……..(ii)Using (i) and (ii): ⇒a20-a=64⇒a2-20a+64=0⇒a2-16a-4a+64=0⇒aa-16-4a-16=0⇒a-16a-4=0⇒a=4, 16If a=4, then b=16.And, if a=16, then b=4.

Page 20.49 (Very Short Answers)

Q4.

Answer :

Let us take a G.P. whose first term is a and common difference is r.
∴ S∞=a1-r ⇒a1-r= 3 …….iAnd, sum of the terms of the G.P. a2, ar2, ar22, … ∞:S’∞= a21-r2 ⇒a21-r2 =92 …….ii⇒2a2 = 9 1-r2 ⇒231-r2 = 9 – 9r2 From i⇒181+r2-2r = 9-9r2⇒18-9 + 18r2+9r2 -36r = 0⇒27r2-36r+9 = 0⇒39r2-12r+3= 0⇒ 9r2-12r+3= 0⇒9r2 -9r-3r+3= 0⇒9rr-1-3r-1=0⇒9r-3r-1=0⇒r=13 and r = 1.But, r = 1 is not possible.∴ r = 13Now, putting r =13 in a1-r = 3:a = 31-13⇒a = 3×23 = 2

Q5.

Answer :

Let us take a G.P. whose first term is A and common ratio is R.
According to the question, we have:ARp-1 = xARq-1 = yARr-1 = z∴ xq-r yr-p zp-q= Aq-r× Rp-1q-r× Ar-p × Rq-1r-p × Ap-q × Rr-1p-q= Aq-r+r-p+p-q × Rpr-pr-q+r+rq-r+p-pq+pr-p-qr+q = A0 × R0=1∴ xq-r yr-p zp-q = 1

Q6.

Answer :

It is given that A1 and A2 are the A.M.s between a and b.Thus, a , A1, A2 and b are in A.P. with common difference d.Here, d = b-a3∴ A1 = a + b-a3 = 2a+b3and A2 = a + 2b-a3=a+2b3It is also given that G1 and G2 are the G.M.s between a and b.Thus, a , G1, G2 and b are in G.P. with common ratio r.Here, r = ba13∴ G1 = aba13 = b13 a13 and G2 = aba13 2= b13 a13 ⇒A1 + A2G1 G2 = 2a+b3+a+2b3 b13 a13 × b13 a13 = a+bab

Q7.

Answer :

Here, second term, a2 = a + dThird term, a3 = a + 2dSixth term, a6 = a + 5d As, a2, a3 and a6 are in G.P.∴ First term of G.P. = a2 = A = a + dSecond term of G.P. = Ar = a + 2dThird term of G.P. =Ar2= a + 5d ∴ a + 2d2= a + d× a + 5d ⇒a2+4ad+4d2=a2+6ad+5d2⇒2ad+d2=0⇒d(2a+d)=0⇒d=0 or 2a+d=0But, d=0 is not possible.∴ d=-2a∴r= a + 2da + d⇒r=a+2(-2a)a+(-2a)⇒r=31=3

Q8.

Answer :

Let the roots of the required quadratic equation be a and b.∴ A = a+b2 and G =abThe equation having a and b as its roots isx2-xa+b+ab = 0⇒x2 -2Ax + G2 = 0 ∵ A =a+b2 and G =ab

Q9.

Answer :

Let G1, G2, …, Gn be n geometric means between two quantities a and b.Thus, a, G1, G2, …, Gn, b is a G.P. Let r be the common ratio of this G.P.∴ r = ba1n+1And, G1 = ar, G2 = ar2, G3 = ar3, …, Gn= arnNow, product of n geometric means = G1·G2·G3· … ·Gn = arar2ar3 … arn=arar2ar3……arn =anr1+2+3 + …+ n=an rnn+12 =anba1n+1nn+12 =anban2=an2bn2 =abn2

Q10.

Answer :

Here, a=1, b, b2, b3, … ∞ form an infinite G.P. ∴ S∞=a =1+b+b2+b3+…∞ = 11-b⇒a = 11-b⇒1-b = 1a ⇒b = 1 – 1a∴ b = a-1a

Page 20.49 (Multiple Choice Questions)

Q1.

Answer :

(b) 111

Let the first term of the G.P. be a.
Let its common ratio be r.
According to the question, we have:
First term = 10 [Sum of all successive terms]
a = 10ar1-r⇒a – ar = 10ar⇒11ar = a⇒r = a11a=111

Q2.

Answer :

(a) −25

If the first term is 1, then, the G.P. will be 1, r, r2, r3, …

Now, 5r2+4r=5r2+45r=5r2+45r+425-425=5r+252-45This will be the least when r+25=0, i.e. r=-25.

Q3.

Answer :

(b) 1

a, b and c are in A.P.∴ 2b = a + c ……..iAnd, x, y and z are in G.P. ∴ y2 = xzNow, xb-c yc-a za-b = xb + a – 2b y2b-a-a za-b From i=xa-b y2b-aza-b=xza-b xzb-a From ii, y2 = xz=xz0 = 1

Q4.

Answer :

(d) 8

The first and the last numbers are equal.Let the four given numbers be p, q, r and p.The first three of four given numbers are in G.P.∴ q2 = p·r ……..iAnd, the last three numbers are in A.P. with common difference 6.We have: First term =qSecond term=r= q+6Third term=p= q + 12Also, 2r = q + pNow, putting the values of p and r in i:q2 = q+12q+6⇒ q2 = q2 +18q +72⇒ 18q +72 = 0⇒ q + 4 = 0⇒q = -4Now, putting the value of q in p = q + 12:p = -4 + 12 = 8

Q5.

Answer :

(a) AP

a, b and c are in G.P. ∴ b2 = acTaking log on both the sides:2log b = log a + log c ……..iNow, a1x = b1y =c1zTaking log on both the sides:log ax = log by = log cz ……..iiNow, comparing i and ii:log ax=log a + log c 2y=log cz⇒log ax=log a + log c 2y and log ax=log cz⇒log a 2y-x=xlog c and log alog c=xz⇒log alog c=x2y-x and log alog c=xz⇒x2y-x=xz ⇒2y = x + zThus, x, y and z are in A.P.

Q6.

Answer :

(d) SRn

Sum of n terms of the G.P., S = arn-1r-1Product of n terms of the G.P., P = anrnn-12Sum of the reciprocals of n terms of the G.P., R = 1rn-1a1r -1=rn-1arn-1r-1∴ P2 = a2r2n-12n ⇒P2 = arn-1r-1rn-1arn-1r-1n ⇒P2 = SRn
Let the first term of the G.P. be a and the common ratio be r.Sum of n terms, S =arn-1r-1Product of the G.P., P=anrnn+12Sum of the reciprocals of n terms, R = 1rn – 1 a1r – 1 = 1-rnrna1-rr p2 = a2rn+12n p2 = arn-1r-11-rnrna1-rrn =SRn

Q7.

Answer :

2.357¯ = 2.0 + 0.357 + 0.000357 + 0.000000357 + … ∞⇒2.357¯ = 2 + 357103+357106+357109+ … ∞⇒2.357¯ = 2 + 3571031-1103⇒2.357¯ =2 +357999⇒2.357¯ = 2355999

Page 20.50 (Multiple Choice Questions)

Q8.

Answer :

(b) q-rp-q

Let a be the first term and d be the common difference of the given A.P.
Then, we have:
pth term, ap= a + p-1dqth term, aq= a + q-1drth term, ar = a + r-1dNow, according to the question the pth, the qth and the rth terms are in G.P.∴ a + q-1d2= a + p-1d×a + r-1d⇒a2+2a q-1d+ q-1d2=a2+adr-1+p-1+p-1 r-1d2⇒ad2q-2-r-p+2+d2q2-2q+1-pr+p+r-1=0⇒a2q-r-p+dq2-2q-pr+p+r=0 ∵ d cannot be 0⇒a=-q2-2q-pr+p+rd2q-r-p∴ Common ratio, r =aq ap= a + q-1da + p-1d=q2-2q-pr+p+rdp+r-2q+ q-1dq2-2q-pr+p+rdp+r-2q+p-1d=q2-2q-pr+p+r+pq+rq-2q2-p-r+2qq2-2q-pr+p+r+p2+pr-2pq-p-r+2q=pq-pr-q2+qrp2+q2-2pq=pq-r-qq-rp-q2=p-qq-rp-q2=q-rp-q

Q9.

Answer :

(b) 3913×919×9127× … ∞=913+19+127+ … ∞Here, it is a G.P. with a=13 and r=13.∴ 9131-13=912=3

Q10.

Answer :

(a) 1/2

Let the G.P. be a, ar, ar2, ar3, …, ∞. S∞=4⇒a1-r=4 (i)Also, sum of the cubes, S1=92⇒a31-r3=92 (ii)Putting the value of a from (i) to (ii):⇒4(1-r)31-r3=92⇒64(1-r)31-r3=92⇒1-r31-r1+r+r2=9264⇒1-r21+r+r2=2316⇒161-2r+r2=231+r+r2⇒7r2+55r+7=0Using the quadratic formula:⇒r=-55+552-4×7×72×7⇒r=-55+552-14214⇒r=-55+282914

Q11.

Answer :

(d) 3/4

Let the terms of the G.P. be a, a2, a3, a4, a5, …, ∞.And, let the common ratio be r.Now, a+a2=1∴ a+ar=1 ……..(i)Also, a=2a2+a3+a4+a5+…∞⇒a=2ar+ar2+ar3+ar4+…∞⇒a=2ar1-r⇒1-r=2r⇒3r=1⇒r=13Putting the value of r in (i):a+a3=1⇒4a3=1⇒4a=3⇒a=34

Q12.

Answer :

an=128, Sn=225 and r=2an=128∴ arn-1=128⇒2n-1a=128⇒2na2=128⇒ 2n=256 a ……..(i)Also, Sn=225⇒arn-1r-1=225⇒a2n-12-1=225⇒a256 a-1=225 Using (i)⇒256-a=225⇒a=256-225⇒a=31

Q13.

Answer :

(d) 4

a2=2 ∴ ar=2 ……..(i)Also, S∞=8⇒a1-r=8⇒a1-2a=8 Using (i)⇒a2=8a-2⇒a2-8a+16=0⇒a-42=0⇒a=4

Q14.

Answer :

(d) 1x+1y=2b

a, b and c are in G.P.∴ b2=ac ……..(i)a, x and b are in A.P.∴ 2x=a+b ……..(ii)Also, b, y and c are in A.P.∴ 2y=b+c ⇒2y=b+b2a Using (i)⇒2y=b+b22x-b Using (ii)⇒2y=b2x-b+b22x-b⇒2y=2bx-b2+b22x-b⇒2y=2bx2x-b⇒y=bx2x-b⇒y2x-b=bx⇒2xy-by=bx⇒bx+by=2xyDividing both the sides by xy:⇒1y+1x=2b

Q15.

Answer :

(a) p3+q3pq

Let the two positive numbers be a and b. a, A and b are in A.P.∴ 2A=a+b (i)Also, a, p, q and b are in G.P.∴ r=ba13Again, p=ar and q=ar2. (ii)Now, 2A=a+b From (i)=a+aba=a+aba133=a+ar3=ar2ar2+ar22ar=p2q+q2p Using (ii)=p3+q3pq

Q16.

Answer :

(a) (2p − q) (p − 2q)

Let the two numbers be a and b.a, p, q and b are in A.P.∴ p-a=q-p=b-q ⇒p-a=q-p and q-p=b-q⇒a=2p-q and b= 2q-p (i)Also, a, G and b are in G.P.∴ G2=ab⇒G2=2p-q2q-p

Q17.

Answer :

(a) 12

Let S=11+x-1-x1+x2+1-x21+x3-1-x31+x4+ … ∞It is clear that it is a G.P. with a=11+x and r=-1-x1+x.∴ S=a1-r⇒S=11+x1–1-x1+x⇒S=11+x1+1-x1+x⇒S=11+x1+x+1-x1+x⇒S=12

Q18.

Answer :

(b) 8

434649412…43x=0.0625-54⇒43+6+9+12+ … +3x=62510000-54⇒431+2+3+4+ … +x=116-54⇒43xx+12=116-54⇒43xx+12=4-2-54Comparing both the sides:⇒3xx+12=108⇒xx+1=72⇒x2+x-72=0⇒x2+9x-8x-72=0⇒xx+9-8x+9=0⇒x+9x-8=0⇒x=8, -9⇒x=8 [∵ x is positive]

Q19.

Answer :

(b) x + 1

∑n=1∞xx+1n-1=1+xx+1+xx+12+xx+13+xx+14+ … ∞=11-xx+1 ∵ it is a G.P. with a =1 and r=xx+1=x+1x+1-x=x+11=x+1

Q20.

Answer :

Let there be 2n terms in a G.P.Let a be the first term and r be the common ratio.∵ S2n=5Sodd terms⇒ar2n-1r-1=5a+ar2+ar4+ar6+ … ar2n-1⇒ar2n-1r-1=5ar2n-1r2-1⇒r2n-1r-1=5r2n-1r2-1⇒rn2-12r-1=5rn2-12r2-1⇒rn-1rn+1r-1=5rn-1rn+1r-1r+1⇒rn-1rn+1r-1r+1-5r-1rn-1rn+1=0⇒rn-1rn+1r-1r+1-5=0But, r =1 or -1 is not possible.∴ r=4

Page 20.51 (Multiple Choice Questions)

Q21.

Answer :

(b) 2

Let the two numbers be a and b.a, x and b are in A.P.∴ 2x=a+b (i)Also, a, y, z and b are in G.P.∴ ya=zy=bz⇒y2=az , yz=ab, z2=by (ii)Now, y3+z3xyz=y2xz+z2xy =1xy2z+z2y=1xazz+byy Using (ii)=1xa+b=2a+ba+b Using (i)=2

Q22.

Answer :

(a) 64

32×3216×32136× … ∞=321+16+136+ … ∞= 3211-16 [∵ it is a G.P.]=3265=2565=26=64

Q23.

Answer :

(b) 4 and 16

Let the two G.M.s between 1 and 64 be G1 and G2.Thus, 1, G1, G2 and 64 are in G.P. 64=1×r3⇒r=643⇒r=4⇒G1=ar=1×4=4And, G2=ar2=1×42=16Thus, 4 and 16 are the required G.M.s.

Q24.

Answer :

Here, am+n=p⇒arm+n-1=p …….(i)Also, am-n=q⇒arm-n-1=q …….(ii)Mutliplying (i) and (ii):⇒arm+n-1arm-n-1=pq⇒a2r2m-2=pq⇒arm-12=pq⇒arm-1=pq⇒am=pqThus, the mth term is pq.

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