# Class 11 Maths Ncert Solutions Ex 9.3

## Class 11 Maths Ncert Solutions Chapter 9 Ex 9.3

Q1. Find 20th and nth term for the G.P 52,54,58,....$\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, . . . .$

Soln:

Given G.P = 52,54,58,....$\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, . . . .$

Here, first term = a = 52$\frac{5}{2}$

Common ratio = r = 5452=12$\frac{ \frac{5}{4} }{ \frac{5}{2}} = \frac{1}{2}$

a20=ar201=52(12)19=5(2)(2)19=5(2)20$a_{20} = ar^{20 – 1} = \frac {5}{2} \left ( \frac{1}{2} \right )^{19} = \frac{5}{(2)(2)^{19}} = \frac{5}{(2)^{20}}$ an=arn1=52(12)n1=5(2)(2)n1=5(2)n$a_{n} = ar^{n – 1} = \frac {5}{2} \left ( \frac{1}{2} \right )^{n – 1} = \frac {5}{(2)(2)^{n – 1}} = \frac{5}{(2)^{n}}$

Q2. Find 12th term for the G.P that has 8th term 192 and common ratio of 2.

Soln:

Given,

Common ratio = r = 2

Assume = first term  = a

a8=ar81=ar7ar7=192a(2)7=(2)6(3)$a_{8} = ar^{8 – 1} = ar^{7} \Rightarrow ar^{7} = 192 a(2)^{7} = (2)^{6} (3)$ a=(2)6×3(2)7=32$\Rightarrow a = \frac{(2)^{6} \times 3}{(2)^{7}} = \frac{3}{2}$ a12=ar121=(32)(2)11=(3)(2)10=3072$a_{12} = ar^{12 – 1} = \left ( \frac{3}{2} \right ) (2)^{11} = (3)(2)^{10} = 3072$

Q3. If p is the 5th, q is the 8th and s is the 11th term of a G.P. Prove that q2 = ps

Soln:

Let the first term be ‘a’ and the common ratio be r for the G.P.

Given condition

a5=ar51=ar4=p....(1)$a_{5} = ar^{5 – 1} = ar^{4} = p . . . . (1)$ a8=ar81=ar7=q....(2)$a_{8} = ar^{8 – 1} = ar^{7} = q . . . . (2)$ a11=ar111=ar10=s....(3)$a_{11} = ar^{11 – 1} = ar^{10} = s . . . . (3)$

Eqn (2) divided by Eqn (1), we have

ar7ar7=qp$\frac{ar^{7}}{ ar^{7}} = \frac{q}{p}$

r3=qp$r^{3} = \frac{q}{p}$  . . . . .  (4)

Eqn (3) divided by Eqn (2), we have

ar10ar7=sq$\frac{ar^{10}}{ ar^{7}} = \frac{s}{q}$

r3=sq$\Rightarrow r^{3} = \frac{s}{q}$  . . . . .  (5)

From eqn (4) and eqn (5), we get

qp=sq$\frac{q}{p} = \frac{s}{q}$ q2=ps$\Rightarrow q^{2} = ps$

Hence proved

Q4. If first term of a G.P is -3 and the 4th term is square of the 2nd term. Find the 7th term.

Soln:

Let the first term be ‘a’ and common ratio be ‘r’.

Therefore, a = -3

We know that, an = arn – 1

a4 = ar3 = (-3)r3

a2 = ar1 = (-3)r

Given condition

(-3)r3 = [(-3) r]2

3r3=9r2r=3a7=ar71=a$\Rightarrow -3r^{3} = 9 r^{2} \Rightarrow r = – 3 a_{7} = ar^{7 – 1} = a$

r6 = (-3) (-3)6 = – (3)7 = -2187

Hence, the 7th term is -2187.

Q5. Which term of

(a) 2,22,4,....is128?$2, 2\sqrt{2}, 4, . . . . \; is \; 128?$

(b) 3,3,33....is729?$\sqrt{3}, 3, 3\sqrt{3} . . . . \; is \; 729?$

(c) 13,19,127,....is119683?$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . . . . \; is \; \frac{1}{19683}?$

Soln:

(a) Given sequence = 2,22,4,....$2, 2\sqrt{2}, 4, . . . .$

Here,

First term = a = 2

Common ratio = r = 222=2$\frac{2\sqrt{2}}{2} = \sqrt{2}$

Assume nth term = 128

an=arn1$a_{n} = a r^{n – 1}$ (2)(2)n1=128$\Rightarrow (2)(\sqrt{2})^{n – 1} = 128$ (2)(2)n12=(2)7$\Rightarrow (2)(2)^{\frac{n – 1}{2}} = (2)^{7}$ (2)n12+1=(2)7$\Rightarrow (2)^{\frac{n – 1}{2} + 1} = (2)^{7}$ n12+1=7$\frac{n – 1}{2} + 1 = 7$ n12=6$⇒ \frac{n – 1}{2} = 6$ n1=12$⇒ n – 1 = 12$ n=13$⇒ n = 13$

Hence, 128 is the 13th term

(b) Given sequence = 3,3,33....$\sqrt{3}, 3, 3\sqrt{3} . . . .$

Here,

First term = a = 3$\sqrt{3}$

Common ratio = r = 33=3$\frac{3}{\sqrt{3}} = \sqrt{3}$

Assume nth term = 729

an=arn1$a_{n} = a r^{n – 1}$ arn1=729$a r^{n – 1} = 729$ (3)(3)n1=729$\Rightarrow (\sqrt{3})(\sqrt{3})^{n – 1} = 729$ (3)12(3)n12=(3)6$\Rightarrow (3)^{\frac{1}{2}}(3)^{\frac{n – 1}{2}} = (3)^{6}$ (3)12+n12=(3)6$\Rightarrow (3)^{\frac{1}{2} + \frac{n – 1}{2}} = (3)^{6}$

Therefore,

12+fracn12=6$\frac{1}{2} + frac{n – 1}{2} = 6$ 1+n12=6$\Rightarrow \frac{1 + n – 1}{2} = 6$ n=12$\Rightarrow n = 12$

Hence, 729 is the 12th term of the sequence.

(c) Given sequence = 13,19,127,....$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . . . .$

Here,

First term = a = 13$\frac{1}{3}$

Common term = r = 19÷13=13$\frac{1}{9} \div \frac{1}{3} = \frac{1}{3}$

Assume the nth term be 119683$\frac{1}{19683}$

an=arn1$a_{n} = a r^{n – 1}$ arn1=119683$a r^{n – 1} = \frac{1}{19683}$ (13)(13)n1=119683$\Rightarrow \left ( \frac{1}{3} \right ) \left ( \frac{1}{3} \right )^{n – 1} = \frac{1}{19683}$ (13)n=(13)9$\Rightarrow \left ( \frac{1}{3} \right )^{n} = \left ( \frac{1}{3} \right )^{9}$

n = 9

Hence, 119683$\frac{1}{19683}$ is the 9th term of the sequence.

Q6. If 27,x,72$\frac{2}{7}, x, -\frac{7}{2}$ are in G.P then find the value of x?

Soln:

Given sequence = 27,x,72$\frac{2}{7}, x, -\frac{7}{2}$

Common ratio = x27=7x2$\frac{x}{\frac{-2}{7}} = \frac{-7x}{2}$

Also, 72x=72x$\frac{\frac{-7}{2}}{x} = \frac{-7}{2x}$

7x2=72x$\frac{-7x}{2} = \frac{-7}{2x}$ x2=2×72×7=1$\Rightarrow x^{2} = \frac{-2 \times 7}{-2 \times 7} = 1$ x=1$\Rightarrow x = \sqrt{1}$ x=±1$\Rightarrow x = \pm 1$

Hence, the sequence is in G.P if x=±1$x = \pm 1$

Q7. The sum of the first 20 terms of the G.P 0.15, 0.015, 0.0015 . . . .

Soln:

Given G.P = 0.15, 0.015, 0.0015 . . . .

Here,

First term = a = 0.15

Common ration = r = 0.0150.15=0.1$\frac{0.015}{0.15} = 0.1$

Sn=a(1rn)1r$S_{n} = \frac{a(1 – r^{n})}{1 – r}$ =0.15[1(0.1)20]10.1$= \frac{0.15[1 – (0.1)^20]}{1 – 0.1}$ =0.150.9[1(0.1)20]$= \frac{0.15}{0.9}[1 – (0.1)^{20}]$ =1590[1(0.1)20]$= \frac{15}{90}[1 – (0.1)^{20}]$ =16[1(0.1)20]$= \frac{1}{6}[1 – (0.1)^{20}]$

Q8. 7,21,37,....$\sqrt{7}, \sqrt{21}, 3\sqrt{7}, . . . .$ is a G.P series. Find sum till the nth term.

Soln:

Given series = 7,21,37,....$\sqrt{7}, \sqrt{21}, 3\sqrt{7}, . . . .$

Here,

First term = a = 7$\sqrt{7}$

Common ratio = r = 217=3$\frac{\sqrt{21}}{7} = \sqrt{3}$

Sn=a(1rn)1r$S_{n} = \frac{a(1 – r^{n})}{1 – r}$ =7[1(3)n]13$= \frac{\sqrt{7}[1 – (\sqrt{3})^{n}]}{1 – \sqrt{3}}$ =7[1(3)n]13×1+31+3$= \frac{\sqrt{7}[1 – (\sqrt{3})^{n}]}{1 – \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}}$ =7(3+1)[1(3)n]13$= \frac{\sqrt{7}(\sqrt{3} + 1)[1 – (\sqrt{3})^{n}]}{1 – 3}$ =7(3+1)[1(3)n]2$= \frac{ – \sqrt{7}(\sqrt{3} + 1)[1 – (\sqrt{3})^{n}]}{2}$

Q9. What is the sum of the nth term of the G.P series

1, -a, a2, -a3 . . . .  (if a ≠ -1)?

Soln:

Given series = 1, -a, a2, -a3 . . .

Here,

The first term of the G.P = a1 = 1

Common ratio of the G.P = r = -a

Sn=a1(1rn)1r$S_{n} = \frac{a_{1}(1 – r^{n})}{1 – r}$ Sn=1[1(a)n]1(a)=[1(a)n]1+a$S_{n} = \frac{1[1 – (-a)^{n}]}{1 – (-a)} = \frac{[1 – (-a)^{n}]}{1 + a}$

Q10. What is the sum of the nth term of the G.P series

x3, x5, x7 . . . . (if x ≠ ± 1)

Soln:

Given series = x3, x5, x7 . . . .

Here,

First term = a = x3

Common ratio = r = x2

Sn=a(1rn)1r=x3[1(x2)n]1x2=x3(1x2n)1x2$S_{n} = \frac{a(1 – r^{n})}{1 – r} = \frac{x^{3}[1 – (x^{2})^{n}]}{1 – x^{2}} = \frac{x^{3}(1 – x^{2n})}{1 – x^{2}}$

Q11. Evaluate 11k=1(2+3k)$\sum_{k = 1}^{11}(2 + 3^{k})$

Soln:

11k=1(2+3k)=11k=1(2)+11k=13k=2(11)+11k=13k=22+11k=13k.....(1)$\sum_{k = 1}^{11}(2 + 3^{k}) = \sum_{k = 1}^{11}(2) + \sum_{k = 1}^{11} 3^{k} = 2(11) + \sum_{k = 1}^{11} 3^{k} = 22 + \sum_{k = 1}^{11} 3^{k} . . . . . (1)$ 11k=13k=31+32+33+....+311$\sum_{k = 1}^{11} 3^{k} = 3^{1} + 3^{2} + 3^{3} + . . . . + 3^{11}$

The above sequence is in a G.P 3, 32, 33, . . .

Sn=a(rn1)r1$S_{n} = \frac{a(r^{n} – 1)}{ r – 1}$ Sn=3[(3)111]31$\Rightarrow S_{n} = \frac{3[(3)^{11} – 1]}{3 – 1}$ Sn=32(3111)$\Rightarrow S_{n} = \frac{3}{2}(3^{11} – 1)$

11k=13k=32(3111)$\sum_{k = 1}^{11} 3^{k} = \frac{3}{2}(3^{11} – 1)$ ….(2)

Substituting eqn (2) in eqn (1), we get

11k=1(2+3k)=22+32(3111)$\sum_{k = 1}^{11} (2 + 3^{k}) = 22 + \frac{3}{2}(3^{11} – 1)$

Q12. Sum of 1st 3 terms in a G.P is 3910$\frac{39}{10}$ and the product of the terms is 1. What is the the terms and common ratio of the sequence?

Soln:

Assume ar,a,ar$\frac{a}{r}, a, ar$ be the 1st 3 terms in the G.P

ar+a+ar=3910$\frac{a}{r} + a + ar = \frac{39}{10}$ . . . . (1)

(ar)(a)(ar)=1$\left (\frac{a}{r} \right ) \left (a \right ) \left ( ar \right ) = 1$ . . . . (2)

From eqn (2), we get a3 = 1

a = 1

Substituting (a = 1) in eqn (1), we get

1r+1+r=3910$\frac{1}{r} + 1 + r = \frac{39}{10}$ 1+r+r2=3910r$\Rightarrow 1 + r + r^{2} = \frac{39}{10}r$ 10+10r+10r239r=0$\Rightarrow 10 + 10 r + 10 r^{2} -39 r = 0$ 10r229r+10=0$\Rightarrow 10 r^{2} – 29 r + 10 = 0$ 10r225r4r+10=0$\Rightarrow 10 r^{2} – 25 r – 4 r + 10 = 0$ ]5r(2r5)2(2r5)=0$]\Rightarrow 5r (2r – 5) -2(2r – 5) = 0$ ](5r2)(2r5)=0$]\Rightarrow (5r – 2) (2r – 5) = 0$ ]r=25or52$]\Rightarrow r = \frac{2}{5} or \frac{5}{2}$

Hence,

The terms are 52,1and25$\frac{5}{2}, 1 \; and \; \frac{2}{5}$

Q13. For the G.P 3, (3)2, (3)3 …. sum of how many terms is 120?

Soln:

Given G.P = 3, (3)2, (3)3 ….

Let n number of terms are required for obtaing the sum of 120

Sn=a(1r)n1r$S_{n} = \frac{a(1 – r)^{n}}{1 – r}$

Here,

First term = a = 3

Common ratio = r = 3

Sn=120=3(3n1)31$S_{n} = 120 = \frac{3(3^{n} – 1)}{3 – 1}$ 120=3(3n1)2$\Rightarrow 120 = \frac{3(3^{n} – 1)}{2}$ 120×23=3n1$\Rightarrow \frac{120 \times 2}{3} = 3^{n} – 1$ 3n1=80$\Rightarrow 3^{n} – 1 = 80$ 3n=81$\Rightarrow 3^{n} = 81$ 3n=34$\Rightarrow 3^{n} = 3^{4}$

n = 4

Hence, 4 terms are required to get the sum of 120

Q14. In a G.P sum of the first 3 terms of are 16 where as sum for the next 3 terms are 128.

Find common ratio, first term and sum of n terms for the G.P

Soln:

Assume G.P = a, ar, a(r)2, a(r)3, . . .

Given condition =

a + a(r)2 + ar + = 16 and a(r)5 + a(r)4 + a(r)3 = 128

a (1 + (r)2 + r) = 16 …… (1)

a(r)3(1+ (r)2 + r) = 128 …….. (2)

eqn (2) divided by eqn (1), we get

ar3(1+r+r2)a(1+r+r2)=12816$\frac{ar^{3}(1 + r + r^{2})}{a(1 + r + r^{2})} = \frac{128}{16}$

r3 = 8

r = 2

substituting r = 2 in eqn(1), we get

a(1 + 2 + 4) = 16

a(7) = 16

a=167$a = \frac{16}{7}$ Sn=167(2n1)21=167(2n1)$S_{n} = \frac{16}{7} \frac{(2^{n} – 1)}{2 – 1} = \frac{16}{7}(2^{n} – 1)$

Q15. Given first term ‘a’ = 729 and the 7th term = 64, then determine S7.

Soln:

Given a = 729 , a7 = 64

Assume common ratio = r

We know that,

an = a rn – 1

a7 = a r7 – 1 = (729) (r)6

64 = 729 (r)6

r6=64729$r^{6} = \frac{64}{729}$ r6=(23)6$r^{6} = \left ( \frac{2}{3}\right )^{6}$ r=23$r = \frac{2}{3}$

Also, we know that

Sn=a(1rn)1r$S_{n} = \frac{a(1 – r^{n})}{1 – r}$ S7=729[1(23)7]123$S_{7} = \frac{729 \left [ 1 – \left ( \frac{2}{3} \right )^{7} \right ] }{1 – \frac{2}{3}}$ =3×729[1(23)7]$= 3 \times 729 \left [ 1 – \left ( \frac{2}{3} \right )^{7} \right ]$ =(3)7[(3)7(2)7(3)7]$= (3)^{7} \left [ \frac{(3)^{7} – (2)^{7}}{(3)^{7}} \right ]$ =(3)7(2)7$= (3)^{7} – (2)^{7}$

= 2187 – 128

= 2059

Q16. The sum of first 2 terms of a G.P is -4 and 5th term is 4 times that of third term. Find the G.P

Soln:

Assume first term = a

Common ratio = r

S2=4=a(1r2)1r$S_{2} = -4 = \frac{a(1 – r^{2})}{1 – r}$ ….. (1)

a5=4×a3$a_{5} = 4 \times a_{3}$ ar4=4ar2r2=4$\Rightarrow ar^{4} = 4ar^{2} \Rightarrow r^{2} = 4$ r=±2$r = \pm 2$

From (1), we get

4=a[1(2)2]12forr=2$-4 = \frac{a\left [ 1 – (2)^{2} \right ]}{1 – 2} \; for \; r = 2$ 4=a(14)1$-4 = \frac{a(1 – 4)}{-1}$

-4 = a(3)

a=43$a = \frac{-4}{3}$

Also, 4=a[1(2)2]1(2)forr=2$-4 = \frac{a\left [ 1 – (-2)^{2} \right ]}{1 – (-2)} \; for \; r = -2$

4=a(14)1+2$-4 = \frac{a(1 – 4)}{1 + 2}$ 4=a(3)3$-4 = \frac{a(- 3)}{3}$

a = 4

Hence, required G.P is 43,83,163or4,8,16,32.$\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3} … \; or \; 4, -8, 16, -32….$

Q17. If p, q, and s are fourth, tenth and sixteenth terms of G.P prove that p, q, s are in G.P

Soln:

Let

First term = a

Common ratio = r

According to given condition,

a4 = ar3 = p ………. (1)

a10 = ar9 = q ………. (2)

a16 = ar15 = x ………. (3)

eqn (2) divided by eqn (1)

qp=ar9ar3qp=r6$\frac{q}{p} = \frac{ar^{9}}{ar^{3}} \Rightarrow \frac{q}{p} = r^{6}$

eqn (3) divided by eqn (2)

sq=ar15ar9sq=r6$\frac{s}{q} = \frac{ar^{15}}{ar^{9}} \Rightarrow \frac{s}{q} = r^{6}$ qp=sq$\frac{q}{p} = \frac{s}{q}$

Q18. 8, 88, 888, 8888, …… are in sequence find sum of the nth terms

Soln:

Given sequence 8, 88, 888, 8888, ……

This sequence isn’t a G.P

It may be changed to a G.P if the terms are written as Sn = 8 + 88 + 888 + 8888 + …..  to n terms

=89[9+99+999+9999+.tonterms]$= \frac{8}{9}\left [ 9 + 99 + 999 + 9999 + ……….\; to \; n \; terms \right ]$ =89[(101)+(1021)+(1031)+(1041)+.tonterms]$= \frac{8}{9}\left [ (10 – 1) + (10^{2} – 1) + (10^{3} – 1) + (10^{4} – 1) + ……….\; to \; n \; terms \right ]$ =89[(10+102+.nterms)(1+1+1.tonterms)]$= \frac{8}{9}\left [ (10 + 10^{2} + …. n \; terms) – (1 + 1 + 1 ……….\; to \; n \; terms) \right ]$ =89[10(10n1)101n]$= \frac{8}{9}\left [ \frac{10(10^{n} – 1)}{10 – 1} – n \right ]$ =89[10(10n1)9n]$= \frac{8}{9}\left [ \frac{10(10^{n} – 1)}{9} – n \right ]$ =8081(10n1)89n$= \frac{80}{81}\left ( 10^{n} – 1 \right ) – \frac{8}{9}n$

Q19. If 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½ are in sequences find the sum of the products

Soln:

Required sum = 2×128+4×32+8×8+16×2+32×12$2 \times 128 + 4 \times 32 + 8\times 8 + 16 \times 2 + 32 \times \frac{1}{2}$

=64[4+2+1+12+122]$= 64 \left [ 4 + 2 + 1 + \frac{1}{2} + \frac{1}{2^{2}} \right ]$

Here, 4,2,1,12,122$4, 2, 1, \frac{1}{2}, \frac{1}{2^{2}}$ is a G.P

Now first term, a = 4

And common ratio, r=12$r = \frac{1}{2}$

We know that,

Sn=a(1rn)1r$S_{n} = \frac{a(1 – r^{n})}{1 – r}$ S5=4[1(12)5]112=4[1132]12=8(32132)=314$S_{5} = \frac{4\left [ 1 – \left ( \frac{1}{2} \right )^{5}\right ]}{1 – \frac{1}{2}} = \frac{4\left [ 1 – \frac{1}{32} \right ]}{ \frac{1}{2}} = 8 \left ( \frac{32 – 1}{32} \right ) = \frac{31}{4}$

Required sum = 64(314)=(16)(31)=496$64\left ( \frac{31}{4} \right ) = (16)(31) = 496$

Q20. Prove that product of corresponding terms from a, ar, ar2,arn-1 and A, AR, AR2,ARn-1 are in sequences find common ratio

Soln:

To show = aA, ar(AR), ar2 (AR2), … arn – 1 (ARn – 1) is a G.P series

SecondtermFirstterm=arARaA=rR$\frac{Second \; term}{First \; term} = \frac{arAR}{aA} = rR$ ThirdtermSecondterm=ar2AR2arAR=rR$\frac{Third \; term}{Second \; term} = \frac{ar^{2}AR^{2}}{arAR} = rR$

Hence, the sequences are in G.P and rR is common ratio.

Q21. If in a G.P of 4 numbers 3rd term is greater to 1st term by 9, and the 2nd term is greater to 4th by 18.

Soln:

Let

First term = a

Common ratio = r

a1 = a,  a2 = ar,  a3 = ar2,  a4 = ar3

From given conditions,

a3 = a1 + 9 => ar2 = a + 9 ………. (1)

a2 = a4 + 18 => ar = ar3 + 18 ………. (2)

From eqn (1) and eqn (2), we get

a(r2 – 1) = 9 ……… (3)

ar(1 – r2) = 18 ……..(4)

Dividing eqn (4) by eqn (3), we get

ar(1r2)a(1r2)=189$\frac{ar(1 – r^{2})}{a( 1 – r^{2})} = \frac{18}{9}$

=>  – r = 2

=> r = -2

Substituting r = -2 in eqn (1), we get

4a = a + 9

=> 3a = 9

=> a = 3

Hence, first four terms of G.P are 3, 3(-2), 3(-2)2, and 3(-2)3

i.e. 3, -6, 12, -24

Q22. If in a G.P a, b and c are the pth, qth and rth terms. Prove that

aq – r.br – p.cp – q = 1

Soln:

Let

First term = a

Common ratio = r

From the given condition

ARp – 1 = a

ARq – 1 = b

ARr – 1 = c

aq – r.br – p.cp – q

=Aqr×R(p1)(qr)×Arp×R(q1)(rp)×Apq×R(r1)(pq)$= A^{q – r} \times R^{(p – 1)(q – r)} \times A^{r – p} \times R^{(q – 1)(r – p)} \times A^{p – q} \times R^{(r – 1)(p – q)}$ =Aqr+rp+pq×R(prprq+r)+(rqr+ppq)+(prpqrq)$= A^{q – r + r – p + p – q} \times R^{(pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr – q)}$

= A0 x R0

= 1

Hence proved

Q23. If in a G.P ‘a’ and ‘b’ are the first and nth term and product of nth terms is P.

Show P2 = (ab)n

Soln:

Given:

First term = a

Last term = b

Therefore,

G.P = a, ar, ar2, ar3 . . . arn – 1, where r is the common ratio.

b = arn – 1…….. (1)

P = product of n terms

= (a)(ar)(ar2)…(arn – 1)

= (a x a x …. a)(r x r2 x … rn – 1)

= an r 1 + 2 + … (n – 1) …… (2)

Here, 1, 2, …..(n – 1) is an A.P

1 + 2 + …. + (n – 1)

=n12[2+(n11)×1]=n12[2+n2]=n(n1)2$= \frac{n – 1}{2}\left [ 2 + (n – 1 – 1) \times 1 \right ] = \frac{n – 1}{2}[2 + n – 2] = \frac{n(n – 1)}{2}$ P=anrn(n1)2$P = a^{n}r^{\frac{n(n -1)}{2}}$ P=a2nrn(n1)$P = a^{2n}r^{n(n -1)}$ =[a2rn1]n$= [a^{2}r^{n – 1}]^{n}$

= (ab)^{n}

Hence proved

Q24. Prove that the ratio of sum of first n terms to sum of terms from (n + 1)th to (2n)th terms is 1/ rn .

Soln:

Let

First term = a

Common ratio = r

Sum of the first n terms = a(1rn)(1r)$\frac{a(1- r^{n})}{(1 – r)}$

Since there are ‘n’ number of terms from (n + 1)th to (2n)th term,

Sum of the terms

Sn=an+1(1rn)1r$S_{n} = \frac{a_{n + 1(1 – r^{n})}}{1 – r}$

Required ratio = a(1rn)(1r)×(1r)arn(1rn)=1rn$\frac{a (1 – r^{n})}{(1 – r)} \times \frac{(1 – r)}{ar^{n}(1 – r^{n})} = \frac{1}{r^{n}}$

Thus, the ratio is 1/rn

Q25. Prove that:

(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2

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

Soln:

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

Therefore

b2 = ac ……………… (2)

c2 = bd ……………… (3)

It is to be proven that,

(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2

R.H.S

= (ab + bc + cd)2

= (ab + ad + cd)2           [From (1)]

= [ab + d(a + c)]2

= a2 (b2) + 2abd(a + c) + (d2)(a + c)2

= (a2) (b2) + 2a2bd + 2acbd + (d2)(a2 + 2ac + c2)

= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2      [From (1) and (2)]

= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2

= a2b2 + a2c2 + a2d2 + b2 x b2 + b2c2 + b2d2 + c2b2 + c2 x c2 + c2d2
[From eqn (1) and (2)]

= a2 (b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2 + c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2) = L.H.S

L.H.S = R.H.S

Q26. Between 3 and 81 insert two numbers such that the sequence is a G.P series.

Soln:

Let the numbers be g1 and g2

So from the given condition 3, g1, g2, 81 is in G.P

Assume first term = a

Common ratio = r

Therefore 81 = (3)(r)3

=> r3 = 27

r = 3 (taking real roots only)

For r = 3,

Q26.

G1 = ar = (3)(3) = 9

G2 = ar2 = (3)(3)2 = 27

Hence, the numbers are 9 and 27

Q27. If a2+1+bn+1an+bn$\frac{a^{2 + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is the geometric mean for a and b. What is the value of n?

Soln:

Mean of a and b is ab$\sqrt{ab}$

From the given condition a2+1+bn+1an+bn=ab$\frac{a^{2 + 1} + b^{n + 1}}{a^{n} + b^{n}} = \sqrt{ab}$

Squaring both the sides, we get

(a2+1+bn+1)2(an+bn)2=ab$\frac{(a^{2 + 1} + b^{n + 1})^{2}}{(a^{n} + b^{n})^{2}} = ab$

=> a2n +1 + 2an + 1 bn + 1 + b2n + 1 = (ab)(a2n + 2anbn + b2n)

=> a2n +1 + 2an + 1 bn + 1 + b2n + 1 = a2n +1b + 2an + 1 bn + 1 + ab2n + 1

=> a2n +1 + b2n + 1 = a2n +1b + ab2n + 1

=> a2n +1 – a2n +1b = ab2n + 1 – b2n + 2

=> a2n +1 (a – b) = b2n + 1 (a – b)

=> (ab)2n+1=1=(ab)0$\left ( \frac{a}{b} \right )^{2n + 1} = 1 = \left ( \frac{a}{b} \right )^{0}$

=> 2n + 1 = 0

=> n=12$n = \frac{-1}{2}$

Q28. Prove that the ratio of two numbers is  (3+22)(322)$(3 + 2\sqrt{2})(3 – 2\sqrt{2})$ if the sum of the two numbers is 6 times their geometric mean.

Soln:

Let the two numbers be a and b.

G.M = ab$\sqrt{ab}$

From given conditions,

a+b=6ab$a + b = 6\sqrt{ab}$ ……. (1)

=> (a + b)2 = 36(ab)

Also,

(a – b)2 = (a + b)2 – 4ab = 36ab – 4ab = 32 ab

=> ab=32ab$a – b = \sqrt{32}\sqrt{ab}$

= 42ab$4\sqrt{2}\sqrt{ab}$ ……. (2)

Eqn(1) + Eqn (2)

2a=(6+42)ab$2a = (6 + 4 \sqrt{2})\sqrt{ab}$

=> a=(3+22)ab$a = (3 + 2 \sqrt{2})\sqrt{ab}$

Putting the value of a in eqn (1),we get

b=6ab(3+22)ab$b = 6\sqrt{ab} – (3 + 2\sqrt{2})\sqrt{ab}$

=> b=(322)ab$b = (3 – 2\sqrt{2})\sqrt{ab}$

ab=(3+22)ab(322)ab=3+22322$\frac{a}{b} = \frac{(3 + 2\sqrt{2})\sqrt{ab}}{(3 – 2\sqrt{2})\sqrt{ab}} = \frac{3 + 2\sqrt{2}}{3 – 2\sqrt{2}}$

Hence, required ratio = (3+22):(322)$\left ( 3 + 2\sqrt{2} \right ) : \left ( 3 – 2\sqrt{2} \right )$

Q29. For two positive numbers if G and A are the G.M and A.M show that the numbers are A±(A+G)(AG)$A \pm \sqrt{(A + G)(A – G)}$

Soln:

Given G and A are A.M and G.M

Assume the numbers be a and b.

AM=A=a+b2(1)$AM = A = \frac{a + b}{2} ……… (1)$ GM=G=ab(2)$GM = G = \sqrt{ab} ……… (2)$

From Eqn(1) and Eqn (2), we get

a + b = 2A ……. (3)

ab = G2 ……. (4)

Putting the value of a and b from (3) and (4)

(a – b)2 = (a + b)2 – 4ab, we get

(a – b)2 = 4A2 – 4G2 = 4(A2 – G2)

(ab)=2(A+G)(AG)(5)$(a – b) = 2\sqrt{(A + G)(A – G)} …… (5)$

From eqn (3) and (5), we get

2a=2A+2(A+G)(AG)$2a = 2A + 2\sqrt{(A + G)(A – G)}$

=> a=A+(A+G)(AG)$a = A + \sqrt{(A + G)(A – G)}$

Putting the value of a in eqn(3), we get

b=2AA(A+G)(AG)=A(A+G)(AG)$b = 2A – A – \sqrt{(A + G)(A – G)} = A – \sqrt{(A + G)(A – G)}$

Hence, the numbers are A±(A+G)(AG)$A \pm \sqrt{(A + G)(A – G)}$

Q30. Bacteria over certain culture multiply twice in one hour. If in initial stage there were only 30 bacteria, what will be the amount of bacteria in the 2nd, 4th and nth hour?

Soln:

Given that amount of bacteria multiplies two times in one hour

Therefore,

Bacteria amount will form G.P

Here, first term = (a) = 30

Common ratio = (r) = 2

Therefore ar2 = a3 = 30 x (2)2 = 120

Therefore amount of bacteria after 2nd hour is 120

ar4 = a5 = 30 x (2)4 = 480

Therefore amount of bacteria after 4th hour is 480

arn x an + 1 = 30 x (2)n

Hence, amount of bacteria after nth hour is 30 x (2)n