NCERT Solutions for Exercise 9.2 Class 11 Maths Chapter 9 - Sequences and Series

Access Sequences And Series Class 11 Chapter 9 Exercise: 9.2

Question:1 Find the sum of odd integers from 1 to 2001.

Answer:

Odd integers from 1 to 2001 are $1,3,5,7...........2001.$

This sequence is an A.P.

Here , first term =a =1

common difference = 2.

We know , $a_n=a+(n-1)d$

$2001=1+(n-1)2$

$\Rightarrow 2000=(n-1)2$

$\Rightarrow 1000=(n-1)$

$\Rightarrow n=1000+1=1001$

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

$=\frac{1001}{2}[2(1)+(1001-1)2]$

$=\frac{1001}{2}[2002]$

$=1001\times 1001$

$=1002001$

The , sum of odd integers from 1 to 2001 is 1002001.

Question:2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Answer:

Numbers divisible by 5 from 100 to 1000 are $105,110,.............995$

This sequence is an A.P.

Here , first term =a =105

common difference = 5.

We know , $a_n=a+(n-1)d$

$995=105+(n-1)5$

$\Rightarrow 890=(n-1)5$

$\Rightarrow 178=(n-1)$

$\Rightarrow n=178+1=179$

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

$=\frac{179}{2}[2(105)+(179-1)5]$

$=\frac{179}{2}[2(105)+178(5)]$

$=179\times 550$

$=98450$

The sum of numbers divisible by 5 from 100 to 1000 is 98450.

Question:3 In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.

Answer:

First term =a=2

Let the series be $2,2+d,2+2d,2+3d,.......................$

Sum of first five terms $=10+10d$

Sum of next five terms $=10+35d$

Given : The sum of the first five terms is one-fourth of the next five terms.

$10+10d=\frac{1}{4}(10+35d)$

$\Rightarrow 40+40d=10+35d$

$\Rightarrow 40-10=35d-40d$

$\Rightarrow 30=-5d$

$\Rightarrow d=-6$

To prove : $\text{Double subscripts: use braces to clarify}$

L.H.S : $\text{Double subscripts: use braces to clarify}$

Hence, 20th term is –112.

Question:4 How many terms of the A.P. $-6,-11/2,-5...$ are needed to give the sum –25?

Answer:

Given : A.P. = $-6,-11/2,-5...$

$a=-6$

$d=\frac{-11}{2}+6=\frac{1}{2}$

Given : sum = -25

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

$\Rightarrow -25=\frac{n}{2}[2(-6)+(n-1)\frac{1}{2}]$

$\Rightarrow -50=n[-12+(n-1)\frac{1}{2}]$

$\Rightarrow -50=-12n+\frac{n^2}{2}-\frac{n}{2}$

$\Rightarrow -100=-24n+n^2-n$

$\Rightarrow n^2-25n+100=0$

$\Rightarrow n^2-5n-20n+100=0$

$\Rightarrow n(n-5)-20(n-5)=0$

$\Rightarrow (n-5)(n-20)=0$

$\Rightarrow n=5or20.$

Question:5 In an A.P., if pth term is 1/q and qth term is 1/p , prove that the sum of first pq terms is 1/2 (pq +1), where $p\ne q$

Answer:

Given : In an A.P., if pth term is 1/q and qth term is 1/p

$a_p=a+(p-1)d=\frac{1}{q}.................(1)$

$a_q=a+(q-1)d=\frac{1}{p}.................(2)$

Subtracting (2) from (1), we get

$\Rightarrow a_p-a_q$

$\Rightarrow (p-1)d-(q-1)d=\frac{1}{q}-\frac{1}{p}$

$\Rightarrow pd-d-qd+d=\frac{p-q}{pq}$

$\Rightarrow (p-q)d=\frac{p-q}{pq}$

$\Rightarrow d=\frac{1}{pq}$

Putting value of d in equation (1),we get

$a+(p-1)\frac{1}{pq}=\frac{1}{q}$

$\Rightarrow a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}$

$\Rightarrow a=\frac{1}{pq}$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

Hence,the sum of first pq terms is 1/2 (pq +1), where $p\ne q$.

Question:6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Answer:

Given : A.P. 25, 22, 19, ….....

$S_n=116$

a=25 , d = -3

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

$\Rightarrow 116=\frac{n}{2}[2(25)+(n-1)(-3)]$

$\Rightarrow 232=n[50-3n+3]$

$\Rightarrow 232=n[53-3n]$

$\Rightarrow 3n^2-53n+232=0$

$\Rightarrow 3n^2-24n-29n+232=0$

$\Rightarrow 3n(n-8)-29(n-8)=0$

$\Rightarrow (3n-29)(n-8)=0$

$\Rightarrow n=8orn=\frac{29}{3}$

n could not be $\frac{29}{3}$ so n=8.

Last term $=a_8=a+(n-1)d$

$=25+(8-1)(-3)$

$=25-21=4$

The, last term of A.P. is 4.

Question:7 Find the sum to n terms of the A.P., whose $k^{th}$ term is 5k + 1.

Answer:

Given : $a_k=5k+1$

$\Rightarrow a+(k-1)d=5k+1$

$\Rightarrow a+kd-d=5k+1$

class 11 maths ch 9 ex 9.Comparing LHS and RHS , we have

$a-d=1$ and $d=5$

Putting value of d,

$a=1+5=6$

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

$S_n=\frac{n}{2}[2(6)+(n-1)5]$

$S_n=\frac{n}{2}[12+5n-5]$

$S_n=\frac{n}{2}[7+5n]$

Question:8 If the sum of n terms of an A.P. is $(pn+q{n}^2)$ , where p and q are constants, find the common difference

Answer:

If the sum of n terms of an A.P. is $(pn+q{n}^2)$,

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

$\Rightarrow \frac{n}{2}[2a+(n-1)d]=pn+q{n}^2$

$\Rightarrow \frac{n}{2}[2a+nd-d]=pn+q{n}^2$

$\Rightarrow an+\frac{n^2}{2}d-\frac{nd}{2}=pn+q{n}^2$

Comparing coefficients of $n^2$ on both side , we get

$\frac{d}{2}=q$

$\Rightarrow d=2q$

The common difference of AP is 2q.

Question:9 The sums of n terms of two arithmetic progressions are in the ratio $5n+4:9n+6$ . Find the ratio of their 18th terms.

Answer:

Given: The sums of n terms of two arithmetic progressions are in the ratio.$5n+4:9n+6$

There are two AP's with first terms =$a_1,a_2$ and common difference = $d_1,d_2$

$\Rightarrow \frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{5n+4}{9n+6}$

$\Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}$

Substituting n=35,we get

$\Rightarrow \frac{2a_1+(35-1)d_1}{2a_2+(35-1)d_2}=\frac{5(35)+4}{9(35)+6}$

$\Rightarrow \frac{2a_1+34d_1}{2a_2+34d_2}=\frac{5(35)+4}{9(35)+6}$

$\Rightarrow \frac{a_1+17d_1}{a_2+17d_2}=\frac{179}{321}$

$\text{Double exponent: use braces to clarify}$

Thus, the ratio of the 18th term of AP's is $179:321$

Question:10 If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

Answer:

Let first term of AP = a and common difference = d.

Then,

$S_p=\frac{p}{2}[2a+(p-1)d]$

$S_q=\frac{q}{2}[2a+(q-1)d]$

Given : $S_p=S_q$

$\Rightarrow \frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d]$

$\Rightarrow p[2a+(p-1)d]=q[2a+(q-1)d]$

$\Rightarrow 2ap+p^{2}d-pd=2aq+q^{2}d-qd$

$\Rightarrow 2ap+p^{2}d-pd-2aq-q^{2}d+qd=0$

$\Rightarrow 2a(p-q)+d(p^2-p-q^2+q)=0$

$\Rightarrow 2a(p-q)+d((p-q)(p+q)-(p-q))=0$

$\Rightarrow 2a(p-q)+d[(p-q)(p+q-1)]=0$

$\Rightarrow (p-q)[2a+d(p+q-1)]=0$

$\Rightarrow 2a+d(p+q-1)=0$

$\Rightarrow d(p+q-1)=-2a$

$\Rightarrow d=\frac{-2a}{p+q-1}$

Now, $S_{(}p+q)=$ $\frac{p+q}{2}[2a+(p+q-1)d]$

$=\frac{p+q}{2}[2a+(p+q-1)\frac{-2a}{p+q-1}]$

$=\frac{p+q}{2}[2a+(-2a)]$

$=\frac{p+q}{2}[0]=0$

Thus, sum of p+q terms of AP is 0.

Question:11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that

$\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Answer:

To prove : $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Let $a_1$ and d be the first term and the common difference of AP, respectively.

According to the given information, we have

$S_p=\frac{p}{2}[2a_1+(p-1)d]=a$

$\Rightarrow [2a_1+(p-1)d]=\frac{2a}{p}............(1)$

$S_q=\frac{q}{2}[2a_1+(q-1)d]=b$

$\Rightarrow [2a_1+(q-1)d]=\frac{2b}{q}............(2)$

$S_r=\frac{r}{2}[2a_1+(r-1)d]=c$

$\Rightarrow [2a_1+(r-1)d]=\frac{2c}{r}............(3)$

Subtracting equation (2) from (1), we have

$\Rightarrow (p-1)d-(q-1)d=\frac{2a}{p}-\frac{2b}{q}$

$\Rightarrow d(p-q-1+1)=\frac{2(aq-bp)}{pq}$

$\Rightarrow d(p-q)=\frac{2(aq-bp)}{pq}$

$\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}$

Subtracting equation (3) from (2), we have

$\Rightarrow (q-1)d-(r-1)d=\frac{2b}{q}-\frac{2c}{r}$

$\Rightarrow d(q-r-1+1)=\frac{2(br-cq)}{qr}$

$\Rightarrow d(q-r)=\frac{2(br-qc)}{qr}$

$\Rightarrow d=\frac{2(br-qc)}{qr(q-r)}$

Equating values of d, we have

$\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}$$=\frac{2(br-qc)}{qr(q-r)}$

$\Rightarrow \frac{2(aq-bp)}{pq(p-q)}$$=\frac{2(br-qc)}{qr(q-r)}$

$\Rightarrow (aq-bp)qr(q-r)=(br-qc)pq(p-q)$

$\Rightarrow (aq-bp)r(q-r)=(br-qc)p(p-q)$

$\Rightarrow (aqr-bpr)(q-r)=(bpr-pqc)(p-q)$

Dividing both sides from pqr, we get

$\Rightarrow (\frac{a}{p}-\frac{b}{q})(q-r)=(\frac{b}{q}-\frac{c}{r})(p-q)$

$\Rightarrow \frac{a}{p}(q-r)-\frac{b}{q}(q-r+p-q)+\frac{c}{r}(p-q)=0$

$\Rightarrow \frac{a}{p}(q-r)-\frac{b}{q}(p-r)+\frac{c}{r}(p-q)=0$

$\Rightarrow \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Hence, the given result is proved.

Question:12 The ratio of the sums of m and n terms of an A.P. is $m^2:n^2$ . Show that the ratio of mth and nth term is $(2m-1):(2n-1)$.

Answer:

Let a and b be the first term and common difference of a AP ,respectively.

Given : The ratio of the sums of m and n terms of an A.P. is $m^2:n^2$ .

To prove : the ratio of mth and nth term is $(2m-1):(2n-1)$.

$\therefore \frac{sumofmterms}{sumofnterms}=\frac{m^2}{n^2}$

$\Rightarrow \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}$

$\Rightarrow \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$

Put $m=2m-1andn=2n-1$, we get

$\Rightarrow \frac{2a+(2m-2)d}{2a+(2n-2)d}=\frac{2m-1}{2n-1}$

$\Rightarrow \frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}.........1$

$\Rightarrow \frac{mthtermofAP}{nthtermofAP}=\frac{a+(m-1)d}{a+(n-1)d}$

From equation (1) ,we get

$\Rightarrow \frac{mthtermofAP}{nthtermofAP}=\frac{2m-1}{2n-1}$

Hence proved.

Question:13 If the sum of n terms of an A.P. is $3n^2+5n$ and its $m^{th}$ term is 164, find the value of m.

Answer:

Given : If the sum of n terms of an A.P. is $3n^2+5n$ and its $m^{th}$ term is 164

Let a and d be first term and common difference of a AP ,respectively.

Sum of n terms = $3n^2+5n$

$\Rightarrow \frac{n}{2}[2a+(n-1)d]=3n^2+5n$

$\Rightarrow 2a+(n-1)d=6n+10$

$\Rightarrow 2a+nd-d=6n+10$

Comparing the coefficients of n on both side , we have

$\Rightarrow d=6$

Also , $2a-d=10$

$\Rightarrow 2a-6=10$

$\Rightarrow 2a=10+6$

$\Rightarrow 2a=16$

$\Rightarrow a=8$

m th term is 164.

$\Rightarrow a+(m-1)d=164$

$\Rightarrow 8+(m-1)6=164$

$\Rightarrow (m-1)6=156$

$\Rightarrow m-1=26$

$\Rightarrow m=26+1=27$

Hence, the value of m is 27.

Question:14 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer:

Let five numbers be A,B,C,D,E.

Then $AP=8,A,B,C,D,E,26$

Here we have,

$a=8,a_7=26,n=7$

$\Rightarrow a+(n-1)d=a_n$

$\Rightarrow 8+(7-1)d=26$

$\Rightarrow 6d=18$

$\Rightarrow d=\frac{18}{6}=3$

Thus, we have $A=a+d=8+3=11$

$B=a+2d=8+(2)3=8+6=14$

$C=a+3d=8+(3)3=8+9=17$

$D=a+4d=8+(4)3=8+12=20$

$E=a+5d=8+(5)3=8+15=23$

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

Question:15 If $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the A.M. between a and b, then find the value of n.

Answer:

Given : $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the A.M. between a and b.

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

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

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

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

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

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

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

$\Rightarrow a^{n-1}=b^{n-1}$

$\Rightarrow {\left[\frac{a}{b}\right]}^{n-1}=1$

$\Rightarrow n-1=0$

$\Rightarrow n=1$

Thus, value of n is 1.

Question:16 Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of $7^{th}$ and $(m-1{)}^{th}$ numbers is 5 : 9. Find the value of m.

Answer:

Let A,B,C.........M be m numbers.

Then, $AP=1,A,B,C..........M,31$

Here we have,

$a=1,a_{m+2}=31,n=m+2$

$\Rightarrow a+(n-1)d=a_n$

$\Rightarrow 1+(m+2-1)d=31$

$\Rightarrow (m+1)d=30$

$\Rightarrow d=\frac{30}{m+1}$

Given : the ratio of $7^{th}$ and $(m-1{)}^{th}$ numbers is 5 : 9.

$\Rightarrow \frac{a+(7)d}{a+(m-1)d}=\frac{5}{9}$

$\Rightarrow \frac{1+7d}{1+(m-1)d}=\frac{5}{9}$

$\Rightarrow 9(1+7d)=5(1+(m-1)d)$

$\Rightarrow 9+63d=5+5md-5d$

Putting value of d from above,

$\Rightarrow 9+63(\frac{30}{m+1})=5+5m\left(\frac{30}{m+1}\right)-5\left(\frac{30}{m+1}\right)$

$\Rightarrow \text{9}(m+1)+1890=5(m+1)+150m-150$

$\Rightarrow \text{9}m+9+1890=5m+5+150m-150$

$\Rightarrow 1890+9-5+150=155m-9m$

$\Rightarrow 2044=146m$

$\Rightarrow m=14$

Thus, value of m is 14.

Question:17 A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment?

Answer:

The first instalment is of Rs. 100.

If the instalment increase by Rs 5 every month, second instalment is Rs.105.

Then , it forms an AP.

$AP=100,105,110,115,.................$

We have ,$a=100andd=5$

$a_n=a+(n-1)d$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

Thus, he will pay Rs. 245 in the 30th instalment.

Question:18 The difference between any two consecutive interior angles of a polygon is $5^{\circ }$. If the smallest angle is ${120}^{\circ }$ , find the number of the sides of the polygon.

Answer:

The angles of polygon forms AP with common difference of $5^{\circ }$ and first term as ${120}^{\circ }$ .

We know that sum of angles of polygon with n sides is $180(n-2)$

$\therefore S_n=180(n-2)$

$\Rightarrow \frac{n}{2}[2a+(n-1)d]=180(n-2)$

$\Rightarrow \frac{n}{2}[2(120)+(n-1)5]=180(n-2)$

$\Rightarrow n[240+5n-5]=360n-720$

$\Rightarrow 235n+5n^2=360n-720$

$\Rightarrow 5n^2-125n+720=0$

$\Rightarrow n^2-25n+144=0$

$\Rightarrow n^2-16n-9n+144=0$

$\Rightarrow n(n-16)-9(n-16)=0$

$\Rightarrow (n-16)(n-9)=0$

$\Rightarrow n=9,16$

Sides of polygon are 9 or 16.