NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 - Sequences and Series

Question 1: Find the $20 ^{th}$ and $n ^{th}$ terms of the G.P. $\frac{5}{2},\frac{5}{4},\frac{5}{8},....$

Answer:

G.P :

$\frac{5}{2},\frac{5}{4},\frac{5}{8},....$

first term = a

$a=\frac{5}{2}$

common ratio =r

$r=\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2}$

$a_n=a.r^{n-1}$

$a_{20}=\frac{5}{2} \cdot\left(\frac{1}{2}\right)^{20-1}$

$a_{20}=\frac{5}{2} \cdot\left(\frac{1}{2^{19}}\right)$

$a_{20}=\frac{5}{2^{20}}$

$a_n=a.r^{n-1}$

$a_n=\frac{5}{2}.\left ( \frac{1}{2} \right )^{n-1}$

$a_n=\frac{5}{2}. \frac{1}{2^{n-1}}$

$a_n=\frac{5}{2^n}$ the nth term of G.P

Question 2: Find the $12 ^{th}$ term of a G.P. whose $8 ^{th}$ term is 192 and the common ratio is 2.

Answer:

First term = a

common ratio =r=2

$8 ^{th}$ term is 192

$a_n=a.r^{n-1}$

$a_8=a.(2)^{8-1}$

$192=a.(2)^{7}$

$a=\frac{2^6.3}{2^7}$

$a=\frac{3}{2}$

$a_n=a.r^{n-1}$

$a_{12}=\frac{3}{2} \cdot(2)^{12-1}$

$a_{12}=\frac{3}{2} \cdot(2)^{11}$

$a_{12}=3 \cdot(2)^{10}$

$a_{12}=3072$ is the $12 ^{th}$ term of a G.P.

Question 3: The $5 ^{th} , 8 ^{th} \: \:and \: \: 11 ^{th}$ terms of a G.P. are p, q and s, respectively. Show that $q ^2 = ps$

Answer:

To prove : $q ^2 = ps$

Let first term=a and common ratio = r

$a_5=a.r^4=p..................(1)$

$a_8=a.r^7=q..................(2)$

$$
a_{11}=a \cdot r^{10}=s .
$$

Dividing equation 2 by 1, we have

$\frac{a.r^7}{a.r^4}=\frac{q}{p}$

$\Rightarrow r^3=\frac{q}{p}$

Dividing equation 3 by 2, we have

$\frac{a \cdot r^{10}}{a \cdot r^7}=\frac{s}{q}$

$\Rightarrow r^3=\frac{s}{q}$

Equating values of $r^3$ , we have

$\frac{q}{p}=\frac{s}{q}$

$\Rightarrow q^2=ps$

Hence proved

Question 4: The $4^{th}$ term of a G.P. is square of its second term, and the first term is -3. Determine its $7^{th}$ term.

Answer:

First term =a= -3

$4^{th}$ term of a G.P. is square of its second term

$\Rightarrow a_4=(a_2)^2$

$\Rightarrow a.r^{4-1}=(a.r^{2-1})^2$

$\Rightarrow a.r^{3}=a^2.r^{2}$

$\Rightarrow r=a=-3$

$a_7=a.r^{7-1}$

$\Rightarrow a_7=(-3).(-3)^{6}$

$\Rightarrow a_7=(-3)^{7}=-2187$

Thus, seventh term is -2187.

Question 5:(a) Which term of the following sequences: $2,2\sqrt 2 , 4 .,....is \: \: 128 ?$

Answer:

Given : $GP = 2,2\sqrt 2 , 4 .,............$

$a=2\, \, \, \, \, and \, \, \, \, \, r=\frac{2\sqrt{2}}{2}=\sqrt{2}$

n th term is given as 128.

$a_n=a.r^{n-1}$

$\Rightarrow 128=2.(\sqrt{2})^{n-1}$

$\Rightarrow 64=(\sqrt{2})^{n-1}$

$\Rightarrow 2^6=(\sqrt{2})^{n-1}$

$\Rightarrow \sqrt{2}^{12}=(\sqrt{2})^{n-1}$

$\Rightarrow n-1=12$

$\Rightarrow n=12+1=13$

The, 13 th term is 128.

Question 5:(b) Which term of the following sequences: $\sqrt 3 ,3 , 3 \sqrt 3 ,...is \: \: 729 ?$

Answer:

Given : $GP=\sqrt 3 ,3 , 3 \sqrt 3 ,........$

$a=\sqrt{3}\, \, \, \, \, and \, \, \, \, \, r=\frac{3}{\sqrt{3}}=\sqrt{3}$

n th term is given as 729.

$a_n=a.r^{n-1}$

$\Rightarrow 729=\sqrt{3}.(\sqrt{3})^{n-1}$

$\Rightarrow 729=(\sqrt{3})^{n}$

$\Rightarrow(\sqrt{3})^{12}=(\sqrt{3})^n$

$\Rightarrow n=12$

The, 12 th term is 729.

Question 5:(c) Which term of the following sequences: $\frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,....is \: \: \frac{1}{19683}?$

Answer:

Given : $GP=\frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,............$

$a=\frac{1}{3}\, \, \, \, \, and \, \, \, \, \, r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}$

n th term is given as $\frac{1}{19683}$

$a_n=a.r^{n-1}$

$\Rightarrow \frac{1}{19683}=\frac{1}{3}.(\frac{1}{3})^{n-1}$

$\Rightarrow \frac{1}{19683}=\frac{1}{3^n}$

$\Rightarrow \frac{1}{3^9}=\frac{1}{3^n}$

$\Rightarrow n=9$

Thus, n=9.

Question 6: For what values of x, the numbers $-\frac{2}{7} ,x, -\frac{7}{2}$ are in G.P.?

Answer:

$GP=-\frac{2}{7} ,x, -\frac{7}{2}$

Common ratio=r.

$r=\frac{x}{\frac{-2}{7}}=\frac{\frac{-7}{2}}{x}$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1$

Thus, for $x=\pm 1$ ,given numbers will be in GP.

Question 7: Find the sum to indicated number of terms in each of the geometric progressions in 0.15, 0.015, 0.0015, ... 20 terms.

Answer:

geometric progressions is 0.15, 0.015, 0.0015, ... .....

a=0.15 , r = 0.1 , n=20

$S_n=\frac{a(1-r^n)}{1-r}$

$S_{20}=\frac{0.15\left(1-(0.1)^{20}\right)}{1-0.1}$

$S_{20}=\frac{0.15\left(1-(0.1)^{20}\right)}{0.9}$

$S_{20}=\frac{0.15}{0.9}\left(1-(0.1)^{20}\right)$

$S_{20}=\frac{15}{90}\left(1-(0.1)^{20}\right)$

$S_{20}=\frac{1}{6}\left(1-0.1^{20}\right)$

Question 8: Find the sum to indicated number of terms in each of the geometric progressions in $\sqrt 7 , \sqrt {21} , 3 \sqrt 7 ,.... n \: \: terms$

Answer:

$GP=\sqrt 7 , \sqrt {21} , 3 \sqrt 7 ,...............$

$a=\sqrt{7}\, \, \, \, and\, \, \, \, \, r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{1-\sqrt{3}}$

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}$

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{1-3} (1+\sqrt{3})$

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{-2} (1+\sqrt{3})$

$S_n=\frac{\sqrt{7}(1+\sqrt{3})}{2} (\sqrt{3}^n-1)$

Question 9: Find the sum to indicated number of terms in each of the geometric progressions in $-a , a^2 , - a ^3 , ... n terms ( if a \neq -1)$

Answer:

The sum to the indicated number of terms in each of the geometric progressions is:

$GP=1,-a , a^2 , - a ^3 , .............$

$a=1\, \, \, and\, \, \, \, r=-a$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{1(1-(-a)^n)}{1-(-a)}$

$S_n=\frac{1(1-(-a)^n)}{1+a}$

$S_n=\frac{1-(-a)^n}{1+a}$

Question 10: Find the sum to indicated number of terms in each of the geometric progressions in $x ^3 , x^5 , x^7 ... n \: \: terms (if \: \: x \neq \pm 1)$

Answer:

$a=x^3\, \, \, and\, \, r=\frac{x^5}{x^3}=x^2$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{x^3(1-(x^2)^n)}{1-x^2}$

$S_n=\frac{x^3\left(1-x^{2 n}\right)}{1-x^2}$

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

Answer:

Given : $\sum_{k = 1}^{11} ( 2+ 3 ^k )=\sum _{k=1}^{11}2 +\sum _{k=1}^{11} 3^k$

$=22 +\sum _{k=1}^{11} 3^k...............(1)$

$\sum_{k=1}^{11} 3^k=3^1+3^2+3^3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots 3^{11}$

These terms form GP with a=3 and r=3.

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{3\left(1-3^{11}\right)}{1-3}$

$S_n=\frac{3\left(1-3^{11}\right)}{-2}$

$S_n=\frac{3\left(3^{11}-1\right)}{2}$$=\sum _{k=1}^{11} 3^k$

$\sum_{k = 1}^{11} ( 2+ 3 ^k )$$=22+\frac{3\left(3^{11}-1\right)}{2}$

Question 12: The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1. Find the common ratio and the terms.

Answer:

Given : The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1.

Let three terms be $\frac{a}{r},a,ar$.

$S_n=\frac{a(1-r^n)}{1-r}$

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

Product of 3 terms is 1.

$\frac{a}{r}\times a\times ar=1$

$\Rightarrow a^3=1$

$\Rightarrow a=1$

Put value of a in equation 1,

$\frac{1}{r}+1+r=\frac{39}{10}$

The three terms of AP are $\frac{5}{2},1,\frac{2}{5}$.

Question 13: How many terms of G.P. $3 , 3 ^ 2 , 3 ^ 3$, … are needed to give the sum 120?

Answer:

G.P.= $3 , 3 ^ 2 , 3 ^ 3$, …............

Sum =120

These terms are GP with a=3 and r=3.

Hence, we have value of n as 4 to get sum of 120.

Question 14: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Answer:

Let GP be $a,ar,ar^2,ar^3,ar^4,ar^5,ar^6................................$

Given: The sum of the first three terms of a G.P. is 16

$a+ar+ar^2=16$

$\Rightarrow a(1+r+r^2)=16...............................(1)$

Given : the sum of the next three terms is128.

$ar^3+ar^4+ar^5=128$

$\Rightarrow ar^3(1+r+r^2)=128...............................(2)$

Dividing equation (2) by (1), we have

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

$\Rightarrow r^3=8$

$\Rightarrow r^3=2^3$

$\Rightarrow r=2$

Putting value of r =2 in equation 1,we have

$\Rightarrow a(1+2+2^2)=16$

$\Rightarrow a(7)=16$

$\Rightarrow a=\frac{16}{7}$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{\frac{16}{7}(1-2^n)}{1-2}$

$S_n=\frac{16}{7}(2^n-1)$

Question 15: Given a G.P. with a = 729 and $7 ^{th}$ term 64, determine $s_7$

Answer:

Given a G.P. with a = 729 and $7 ^{th}$ term 64.

$a_n=a.r^{n-1}$

$\Rightarrow 64=729.r^{7-1}$

$\Rightarrow r^6=\frac{64}{729}$

$\Rightarrow r^6=\left ( \frac{2}{3} \right )^6$

$\Rightarrow r=\frac{2}{3}$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_7=\frac{729(1-\left ( \frac{2}{3} \right )^7)}{1-\frac{2}{3}}$

$S_7=\frac{729(1-\left ( \frac{2}{3} \right )^7)}{\frac{1}{3}}$

$S_7=3\times 729 \left ( \frac{3^7-2^7}{3^7} \right )$

$S_7= \left ( 3^7-2^7 \right )$

$S_7= 2187-128$

$S_7= 2059$(Answer)

Question 16: Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

Answer:

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term

Let first term be a and common ratio be r

$a_5=4.a_3$

$\Rightarrow a.r^{5-1}=4.a.r^{3-1}$

$\Rightarrow a.r^{4}=4.a.r^{2}$

$\Rightarrow r^{2}=4$

$\Rightarrow r=\pm 2$

If r=2, then

$S_n=\frac{a(1-r^n)}{1-r}$

$\Rightarrow \frac{a(1-2^2)}{1-2}=-4$

$\Rightarrow \frac{a(1-4)}{-1}=-4$

$\Rightarrow a(-3)=4$

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

If r= - 2, then

$S_n=\frac{a(1-r^n)}{1-r}$

$\Rightarrow \frac{a(1-(-2)^2)}{1-(-2)}=-4$

$\Rightarrow \frac{a(1-4)}{3}=-4$

$\Rightarrow a(-3)=-12$

$\Rightarrow a=\frac{-12}{-3}=4$

Thus, required GP is $\frac{-4}{3},\frac{-8}{3},\frac{-16}{3},.........$ or $4,-8,-16,-32,..........$

Question 17: If the $4 ^{th} , 10 ^{th} , 16 ^ {th}$ terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.

Answer:

Let x,y, z are in G.P.

Let first term=a and common ratio = r

$a_4=a.r^3=x..................(1)$

$
a_{10}=a \cdot r^9=y .
$

$
a_{16}=a . r^{15}=z .
$

Dividing equation 2 by 1, we have

$\frac{a.r^9}{a.r^3}=\frac{y}{x}$

$\Rightarrow r^4=\frac{y}{x}$

Dividing equation 3 by 2, we have

$\frac{a \cdot r^{15}}{a \cdot r^9}=\frac{z}{y}$

$\Rightarrow r^4=\frac{z}{y}$

Equating values of $r^4$ , we have

$\frac{y}{x}=\frac{z}{y}$

Thus, x,y,z are in GP

Question 18: Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Answer:

8, 88, 888, 8888… is not a GP.

It can be changed in GP by writing terms as

$S_n=8+88+888+8888+.............$ to n terms

$S_n=\frac{8}{9}[9+99+999+9999+................]$

$S_n=\frac{8}{9}[(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+................]$

$S_n=\frac{8}{9}[(10+10^2+10^3+........)-(1+1+1.....................)]$

$S_n=\frac{8}{9}[\frac{10(10^n-1)}{10-1}-(n)]$

$S_n=\frac{8}{9}[\frac{10(10^n-1)}{9}-(n)]$

$S_n=\frac{80}{81}(10^n-1)-\frac{8n}{9}$

Question 19: Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,1/2

Answer:

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

$Required \, \, sum=64\left [ 4+2+1+\frac{1}{2}+\frac{1}{2^2} \right ]$

Here, $4,2,1,\frac{1}{2},\frac{1}{2^2}$ is a GP.

first term =a=4

common ratio =r

$r=\frac{1}{2}$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_5=\frac{4(1-\left ( \frac{1}{2} \right )^5)}{1-\frac{1}{2}}$

$S_5=\frac{4(1-\left ( \frac{1}{2} \right )^5)}{\frac{1}{2}}$

$S_5=8(1-\left ( \frac{1}{32} \right ))$

$S_5=8(\frac{31}{32})$

$S_5=\frac{31}{4}$

$Required \, \, sum=64\left [ \frac{31}{4} \right ]$

$Required \, \, sum=16\times 31=496$

Question 20: Show that the products of the corresponding terms of the sequences $a,ar, ar^2 , ...ar^{n-1} \: \: and\: \: A ,AR, AR^2 ....AR^{n-1}$ form a G.P, and find the common ratio.

Answer:

To prove : $aA,arAR,ar^2AR^2,...................$ is a GP.

$\frac{second \, \, term}{first\, \, term}=\frac{arAR}{aA}=rR$

$\frac{third \, \, term}{second\, \, term}=\frac{ar^2AR^2}{arAR}=rR$

Thus, the above sequence is a GP with common ratio of rR.

Question 21: Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the $4 ^{th}$ by 18.

Answer:

Let first term be a and common ratio be r.

$a_1=a,a_2=ar,a_3=ar^2,a_4=ar^3$

Given : the third term is greater than the first term by 9, and the second term is greater than the $4 ^{th}$ by 18.

$a_3=a_1+9$

$\Rightarrow ar^2=a+9$

$\Rightarrow a(r^2-1)=9.................1$

$a_2=a_4+18$

$\Rightarrow ar=ar^3+18$

$\Rightarrow ar(1-r^2)=18......................2$

Dividing equation 2 by 1 , we get

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

$\Rightarrow r=-2$

Putting value of r , we get

$4a=a+9$

$\Rightarrow 4a-a=9$

$\Rightarrow 3a=9$

$\Rightarrow a=3$

Thus, four terms of GP are $3,-6,12,-24.$

Question 22: If the $p^{th} , q ^{th} , r ^{th}$ terms of a G.P. are a, b and c, respectively. Prove that $a ^{ q-r } b ^{r- p } C ^{p-q} = 1$

Answer:

To prove : $a ^{ q-r } b ^{r- p } C ^{p-q} = 1$

Let A be the first term and R be common ratio.

According to the given information, we have

$a_p=A.R^{p-1}=a$

$a_q=A.R^{q-1}=b$

$a_r=A.R^{r-1}=c$

L.H.S : $a ^{ q-r } b ^{r- p } C ^{p-q}$

$=A^{q-r}.R^{(q-r)(p-1)}.A^{r-p}.R^{(r-p)(q-1)}.A^{p-q}.R^{(p-q)(r-1)}$

$=A^{q-r+r-p+p-q}.R^{(qp-rp-q+r)+(rq-pq+p-r)+(pr-p-qr+q)}$

$=A^0.R^0=1$=RHS

Thus, LHS = RHS.

Hence proved.

Question 23: If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that $P^2 = ( ab)^n$.

Answer:

Given : First term =a and n th term = b.

Common ratio = r.

To prove : $P^2 = ( ab)^n$

Then , $GP = a,ar,ar^2,ar^3,ar^4,..........................$

$a_n=a.r^{n-1}=b..................................1$

P = product of n terms

$P=(a).(ar).(ar^2).(ar^3)..............(ar^{n-1})$

$P=(a.a.a...............a)((1).(r).(r^2).(r^3)..............(r^{n-1}))$

$P=(a^n)(r^{1+2+.........(n-1)})........................................2$

Here, $1+2+.........(n-1)$ is a AP.

$\therefore\, \, \, sum= \frac{n}{2}\left [2a+(n-1)d \right ]$

$= \frac{n-1}{2}\left [2(1)+(n-1-1)1 \right ]$

$= \frac{n-1}{2}\left [2+n-2 \right ]$

$= \frac{n-1}{2}\left [n \right ]$

$= \frac{n(n-1)}{2}$

Put in equation (2),

$P=(a^n)(r^{\frac{n(n-1)}{2}})$

$P^2=\left(a^{2 n}\right)\left(r^{n(n-1)}\right)$

$P^2=(a. a.r^{(n-1)})^n$

$P^2=(a.b)^n$

Hence proved.

Question 24: Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $( n+1)^{th} \: \: to\: \: (2n)^{th}$ term is $\frac{1}{r^n}$

Answer:

Let first term =a and common ratio = r.

$sum \, \, of\, \, n\, \, terms=\frac{a(1-r^n)}{1-r}$

Since there are n terms from (n+1) to 2n term.

Sum of terms from (n+1) to 2n.

$S_n=\frac{a_{(n}+{ }_{1)}\left(1-r^n\right)}{1-r}$

$\left.a_{(n}+1\right)=a \cdot r^{n+1-1}=a r^n$

Thus, the required ratio = $\frac{a(1-r^n)}{1-r}\times \frac{1-r}{ar^n(1-r^n)}$

$=\frac{1}{r^n}$

Thus, the common ratio of the sum of first n terms of a G.P. to the sum of terms from $( n+1)^{th} \: \: to\: \: (2n)^{th}$ term is $\frac{1}{r^n}$.

Question 25: If a, b, c and d are in G.P. show that $(a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2 .$

Answer:

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

$bc=ad....................(1)$

$b^2=ac....................(2)$

$c^2=bd....................(3)$

To prove : $(a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2 .$

RHS : $(ab + bc + cd)^2 .$

$=(ab + ad + cd)^2 .$

$=(ab + d (a+ c))^2 .$

$=a^2b^2 + d^2 (a+ c)^2 + 2(ab)(d(a+c))$

$=a^2b^2 + d^2 (a^2+ c^2+2ac) + 2a^2bd+2bcd$

Using equations (1) and (2),

$=a^2b^2 + 2a^2c^2+ 2b^2c^2+d^2a^2+2d^2b^2+d^2c^2$

$=a^2b^2 + a^2c^2+ a^2c^2+b^2c^2+b^2c^2+d^2a^2+d^2b^2+d^2b^2+d^2c^2$

$=a^2b^2 + a^2c^2+ a^2d^2+b^2.b^2+b^2c^2+b^2d^2+c^2b^2+c^2.c^2+d^2c^2$

$=a^2(b^2 + c^2+ d^2)+b^2(b^2+c^2+d^2)+c^2(b^2+c^2+d^2)$

$=(b^2 + c^2+ d^2)(a^2+b^2+c^2)$ = LHS

Hence proved

Question 26: Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Answer:

Let A, B be two numbers between 3 and 81 such that the series 3, A, B,81 forms a GP.

Let a= first term and common ratio =r.

$\therefore a_4=a.r^{4-1}$

$81=3.r^{3}$

$27=r^{3}$

$r=3$

For $r=3$,

$A=ar=(3)(3)=9$

$B=ar^2=(3)(3)^2=27$

The, required numbers are 9,27.

Question 27: Find the value of n so that $\frac{a^{n+1}+ b ^{n+1}}{a^n+b^n}$ may be the geometric mean between a and b.

Answer:

M of a and b is $\sqrt{ab}.$

Given :

$\frac{a^{n+1}+ b ^{n+1}}{a^n+b^n}=\sqrt{ab}$

Squaring both sides ,

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

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

$\Rightarrow\left(a^{2 n+2}+b^{2 n+2}+2 \cdot a^{n+1} \cdot b^{n+1}\right)=\left(a^{2 n}+b^{2 n}+2 \cdot a^n \cdot b^n\right) a b$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2}+2.a^{n+1}.b^{n+1})=({a^{2n+1}.b+a.b^{2n+1}+2.a^{n+1}.b^{n+1}} \right )$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2})=({a^{2n+1}.b+a.b^{2n+1}} \right )$

$\Rightarrow a^{2n+2}-{a^{2n+1}.b=a.b^{2n+1}} )- b ^{2n+2}$

$\Rightarrow a^{2n+1}(a-b)=b^{2n+1}( a-b)$

$\Rightarrow a^{2n+1}=b^{2n+1}$

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

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

$\Rightarrow 2n+1=0$

$\Rightarrow 2n=-1$

$\Rightarrow n=\frac{-1}{2}$

Question 28: The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 )$

Answer:

Let there be two numbers a and b

geometric mean $=\sqrt{ab}$

According to the given condition,

$a+b=6\sqrt{ab}$

$(a+b)^2=36(ab)$.............................................................(1)

Also,$(a-b)^2=(a+b)^2-4ab=36ab-4ab=32ab$

$(a-b)=\sqrt{32}\sqrt{ab}$

$(a-b)=4\sqrt{2}\sqrt{ab}$.......................................................(2)

From (1) and (2), we get

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

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

Putting the value of 'a' in (1),

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

$b=(3-2\sqrt{2})\sqrt{ab}$

$\frac{a}{b}=\frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}$

$\frac{a}{b}=\frac{(3+2\sqrt{2})}{(3-2\sqrt{2})}$

Thus, the ratio is $( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 )$

Question 29: If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $A \pm \sqrt{( A+G)(A-G)}$

Answer:

If A and G be A.M. and G.M., respectively between two positive numbers,
Two numbers be a and b.

$AM=A=\frac{a+b}{2}$

$\Rightarrow a+b=2A$...................................................................1

$GM=G=\sqrt{ab}$

$\Rightarrow ab=G^2$...........................................................................2

We know $(a-b)^2=(a+b)^2-4ab$

Put values from equation 1 and 2,

$(a-b)^2=4A^2-4G^2$

$(a-b)^2=4(A^2-G^2)$

$(a-b)^2=4(A+G)(A-G)$

$(a-b)=4\sqrt{(A+G)(A-G)}$..................................................................3

From 1 and 3 , we have

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

$\Rightarrow a=A+\sqrt{(A+G)(A-G)}$

Put value of a in equation 1, we get

$b=2A-A-\sqrt{(A+G)(A-G)}$

$\Rightarrow b=A-\sqrt{(A+G)(A-G)}$

Thus, numbers are $A \pm \sqrt{( A+G)(A-G)}$

Question 30: The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?

Answer:

The number of bacteria in a certain culture doubles every hour.It forms GP.

Given : a=30 and r=2.

$a_3=a.r^{3-1}=30(2)^2=120$

$a_5=a.r^{5-1}=30(2)^4=480$

$a_n+_1=a.r^{n+1-1}=30(2)^n$

Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and $30(2)^n$ respectively.

Question:31 What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Answer:

Given: Bank pays an annual interest rate of 10% compounded annually.

Rs 500 amounts are deposited in the bank.

At the end of the first year, the amount

$=500\left ( 1+\frac{1}{10} \right )=500(1.1)$

At the end of the second year, the amount $=500(1.1)(1.1)$

At the end of the third year, the amount $=500(1.1)(1.1)(1.1)$

At the end of 10 years, the amount $=500(1.1)(1.1)(1.1)........(10times)$

$=500(1.1)^{10}$

Thus, at the end of 10 years, amount $=Rs. 500(1.1)^{10}$

Question 32: If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation

Answer:

Let roots of the quadratic equation be a and b.

According to given condition,

$AM=\frac{a+b}{2}=8$

$\Rightarrow (a+b)=16$

$GM=\sqrt{ab}=5$

$\Rightarrow ab=25$

We know that $x^2-x(sum\, of\, roots)+(product\, of\, roots)=0$

$x^2-x(16)+(25)=0$

$x^2-16x+25=0$

Thus, the quadratic equation = $x^2-16x+25=0$