NCERT Solutions for Exercise 9.3 Class 11 Maths Chapter 9 - Sequences and Series

# NCERT Solutions for Exercise 9.3 Class 11 Maths Chapter 9 - Sequences and Series

Edited By Vishal kumar | Updated on Nov 08, 2023 10:28 AM IST

## NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.3- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.3-In the previous exercise, you have already learned about the arithmetic progression, nth term and sum of the arithmetic progression, arithmetic mean, etc. In NCERT syllabus Class 11 Maths chapter 9 exercise 9.3, you will solve questions on geometric progression, nth term of the geometric progression, the sum of the geometric progression, geometric mean, relationship between the arithmetic mean and geometric mean, etc. You must have learned about the geometric progression, the nth term of the geometric progression, and the sum of the terms of the geometric progression in the previous classes. As you are familiar with these topics, it won't take much effort to understand these topics.

You will learn some new topics like arithmetic mean, geometric mean and the relationship between the arithmetic mean and geometric mean in the exercise 9.3 Class 11 Maths. Arithmetic mean and geometric mean are very similar to arithmetic progression and geometric progression respectively, so you will understand them very easily. Check NCERT Solutions if you are looking for NCERT solutions for Class 6 to Class 12 at one place. All the solutions are expertly crafted by subject experts at Careers360 in a user-friendly and detailed manner. Furthermore, PDF versions of these solutions are available for free, providing students with convenient access.

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**As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 8.

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

G.P :

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

first term = a

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

common ratio =r

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

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

$a_2_0=\frac{5}{2}.(\frac{1}{2})^{20-1}$

$a_2_0=\frac{5}{2}.(\frac{1}{2^{19}})$

$a_2_0=\frac{5}{2^{20}}$

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

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

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

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

First term = a

common ratio =r=2

$8 ^{th}$ term is 192

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

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

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

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

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

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

$a_1_2=\frac{3}{2}. ( 2 )^{12-1}$

$a_1_2=\frac{3}{2}. ( 2 )^{11}$

$a_1_2= 3. ( 2 )^{10}$

$a_1_2= 3072$ is the $12 ^{th}$ term of a G.P.

To prove : $q ^2 = ps$

Let first term=a and common ratio = r

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

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

$a_1_1=a.r^1^0=s..................(3)$

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

$\frac{a.r^1^0}{a.r^7}=\frac{s}{q}$

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

Equating values of $r^3$ , we have

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

$\Rightarrow q^2=ps$

Hence proved

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.

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.

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

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

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

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

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

$\Rightarrow n-1=12$

$\Rightarrow n=12+1=13$

The, 13 th term is 128.

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.

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

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

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

$\Rightarrow( \sqrt{3})^1^2=(\sqrt{3})^{n}$

$\Rightarrow n=12$

The, 12 th term is 729.

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.

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

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.

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

a=0.15 , r = 0.1 , n=20

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

$S_2_0=\frac{0.15(1-(0.1)^{20})}{1-0.1}$

$S_2_0=\frac{0.15(1-(0.1)^{20})}{0.9}$

$S_2_0=\frac{0.15}{0.9}(1-(0.1)^{20})$

$S_2_0=\frac{15}{90}(1-(0.1)^{20})$

$S_2_0=\frac{1}{6}(1-0.1^{20})$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sum _{k=1}^{11} 3^k=3^1+3^2+3^3+....................3^1^1$

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

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

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

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

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

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

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 $\frac{5}{2},1,\frac{2}{5}$.

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.

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

Given : The sum of 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

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

$\Rightarrow a(7)=16$

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

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

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

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

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)

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 $4,-8,-16,-32,..........$

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

Let first term=a and common ratio = r

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

$a_1_0=a.r^9=y..................(2)$

$a_1_6=a.r^1^5=z..................(3)$

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

$\frac{a.r^1^5}{a.r^9}=\frac{z}{y}$

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

Equating values of $r^4$ , we have

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

Thus, x,y,z are in GP

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

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

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

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

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

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

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

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

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

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

first term =a=4

common ratio =r

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

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

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

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

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

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

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

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

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

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.

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 $3,-6,12,-24.$

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.

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),

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

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

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

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

Hence proved .

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.

$S_n=\frac{a_(_n+_1_)(1-r^n)}{1-r}$

$a_(_n+_1)=a.r^{n+1-1}=ar^n$

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

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 equation (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

Let A, B be two numbers between 3 and 81 such that 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.

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 ,

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

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

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

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

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

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

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

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

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

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

$\Rightarrow 2n+1=0$

$\Rightarrow 2n=-1$

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

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 $( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 )$

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 $A \pm \sqrt{( A+G)(A-G)}$

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.

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 $=Rs. 500(1.1)^{10}$

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

## More About NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3:-

Class 11 Maths chapter 9 exercise 9.3 consists of questions related to finding the nth term of the geometric progression, the sum of the terms of geometric progression, arithmetic mean, geometric mean, etc. The Class 11 Maths chapter 9 exercise 9.3 is quite lengthy as compared to other exercises of this chapter. Also, exercise 9.3 Class 11 Maths is the most important exercise of this chapter as it covers topics such as geometric mean, arithmetic mean.

Also Read| Sequences And Series Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3:-

• NCERT book Class 11 maths chapter 9 exercise 9.3 is more lengthy and many concepts of progression are covered in it.
• Most of the questions in exercise 9.3 Class 11 Maths are conceptually based on the geometric progression and arithmetic progression concepts.
• You must try to solve all the problems of this exercise on your own before moving to the NCERT solutions for Class 11 Maths chapter 9 exercise 9.3.
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### Key Features of NCERT 11th Class Maths Exercise 9.3 Answers

1. Comprehensive Coverage: The answers encompass all the ex 9.3 class 11 problem of the NCERT 11th Class Mathematics textbook.

2. Step-by-Step Solutions: The class 11 maths ex 9.3 solutions are presented in a step-by-step format, making it easier for students to understand and follow the problem-solving process.

3. Clarity and Precision: The class 11 ex 9.3 answers are written with clarity and precision, ensuring that students can grasp the mathematical concepts and methods required for effective problem-solving.

4. Conceptual Understanding: The 11th class maths exercise 9.3 answers aim to promote a deep understanding of mathematical concepts rather than rote memorization, encouraging critical thinking and problem-solving skills.

5. Free Accessibility: Typically, these class 11 maths ex 9.3 are available free of charge, allowing students to access them without any cost, making it a valuable resource for self-study.

6. Supplementary Learning Resource: These ex 9.3 class 11 answers can be used as a supplementary learning resource to reinforce classroom instruction and aid in exam preparation.

7. Homework and Practice: Students can utilize these answers to verify their work, practice problem-solving, and enhance their overall performance in mathematics.

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Find the 5th term of G.P. 2,4,8...

Given a = 2

r = 4/2 = 2

a_5 = ar^4 = 2(2)^4 = 32

2. If the first term and third term of the G.P. are 3, 48 respectively than find the common ratio of the G.P. ?

Given a = 3

ar^2 = 48

3(r^2) = 48

r^2 = 16

Common ratio (r) = 4

3. Find the arithmetic mean of two positive integers a and b ?

A.M. = (a+b)/2

4. Find the geometric mean of two positive integers a and b ?

G.M. = (ab)^(1/2)

5. What is the sum of first n natural numbers?

Sum of first n natural numbers  = n(n+1)/2

6. What is the sum of square of first n natural numbers?

Sum of square of first n natural numbers  = n(n+1)(2n+1)/6

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