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: $2,2\sqrt{2},4.,....is128?$

Answer:

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

$a=2andr=\frac{2\sqrt{2}}{2}=\sqrt{2}$

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: $\sqrt{3},3,3\sqrt{3},...is729?$

Answer:

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

$a=\sqrt{3}andr=\frac{3}{\sqrt{3}}=\sqrt{3}$

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

$a=\frac{1}{3}andr=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}$

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

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

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

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

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

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

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

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

Answer:

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

$a=\sqrt{7}andr=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$

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

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

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

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

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

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

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

Answer:

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

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

$a=1andr=-a$

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

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

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

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

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

Answer:

$a=x^{3}andr=\frac{x^5}{x^3}=x^2$

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

$\sum _{k=1}^{11}(2+3^k)$$=22+\frac{3(3^{11}-1)}{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 $a,ar,a{r}^2,a{r}^3,a{r}^4,a{r}^5,a{r}^6................................$

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

$a+ar+a{r}^2=16$

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

Given : the sum of the next three terms is128.

$a{r}^3+a{r}^4+a{r}^5=128$

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

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

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

$\Rightarrow r^3=8$

$\Rightarrow r^3=2^3$

$\Rightarrow r=2$

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.

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

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

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

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

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

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

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

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

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

$S_7=(3^7-2^7)$

$S_7=2187-128$

$S_7=2059$(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:

$Requiredsum=2\times 128+4\times 32+8\times 8+16\times 2+32\times \frac{1}{2}$

$Requiredsum=64[4+2+1+\frac{1}{2}+\frac{1}{2^2}]$

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

first term =a=4

common ratio =r

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

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

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

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

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

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

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

$Requiredsum=64\left[\frac{31}{4}\right]$

$Requiredsum=16\times 31=496$

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

Answer:

To prove : $aA,arAR,a{r}^{2}A{R}^2,...................$ is a GP.

$\frac{secondterm}{firstterm}=\frac{arAR}{aA}=rR$

$\frac{thirdterm}{secondterm}=\frac{a{r}^{2}A{R}^2}{arAR}=rR$

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.

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

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

$a_3=a_1+9$

$\Rightarrow a{r}^2=a+9$

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

$a_2=a_4+18$

$\Rightarrow ar=a{r}^3+18$

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

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 , $GP=a,ar,a{r}^2,a{r}^3,a{r}^4,..........................$

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

P = product of n terms

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

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

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

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

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

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

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

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

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

Put in equation (2),

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

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

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

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

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.

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