NCERT Solutions for Exercise 9.2 Class 11 Maths Chapter 9

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

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NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2- In the previous exercise, you have already learned about the sequence, and series. The progression is a sequence of a specific pattern that you will learn in the NCERT solutions for Class 11 Maths chapter 9 exercise 9.2. The sequence is called an arithmetic progression if the difference between two consecutive terms is constant. In NCERT book exercise 9.2 Class 11 Maths, you will learn about the arithmetic progression(A. P.), properties of arithmetic progression, the sum of the arithmetic progression, arithmetic mean, etc. Also, you will get questions related to finding the nth term of arithmatic progression, using properties of arithmetic progression to asthmatic mean, etc.

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NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2- Download Free PDF
Download the PDF of NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2
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Question:1 Find the sum of odd integers from 1 to 2001.
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NCERT Solutions for Exercise 9.2 Class 11 Maths Chapter 9 - Sequences and Series

As you have already studied the arithmetic progression, nth term of the arithmetic progression, the sum of the arithmetic progression in the previous classes you won't get any difficulty to understand the Class 11 Maths chapter 9 exercise 9.2. This exercise is very important to solving some real-life problems as well as understanding upcoming exercises of this chapter. NCERT syllabus Exercise 9.2 Class 11 Maths is the basic exercise to understand the arithmatic progression and upcoming concepts like arithmetic mean, geometric mean, the relation between the arithmetic mean and geometric mean etc. Check NCERT Solutions if you are looking for NCERT solutions for Science and Maths of other Classes as well. You will get detailed solutions from Classes 6 to 12 for Science and Maths.

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

Download the PDF of NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2

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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 : $a_2_0=-112$

L.H.S : $a_2_0=a+(20-1)d=2+(19)(-6)=2-114=-112=R.H.S$

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \, \, \, \, n=5\, \, or\, \, 20.$

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 $p \neq q$

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

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

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

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

$\therefore \, \, S_p_q=\frac{pq}{2}[2.\frac{1}{pq}+(pq-1).\frac{1}{pq}]$

$\Rightarrow \, \, S_p_q=\frac{1}{2}[2+(pq-1)]$

$\Rightarrow \, \, S_p_q=\frac{1}{2}[pq+1]$

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \, \, n=8\, \, or\, \, \, n=\frac{29}{3}$

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 $( pn + qn ^ 2 )$ , where p and q are constants, find the common difference

Answer:

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

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

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

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

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

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

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

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

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

$\Rightarrow \, \, \frac{18^t^h \, term \, of\, first \, AP}{18^t^h\, term\, of\, second\, AP}=\frac{179}{321}$

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 : $S_p=S_q$

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

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

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

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

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

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

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

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

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

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

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

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 $( 2m-1) : ( 2n- 1 )$ .

$\therefore \, \frac{sum\, of\, m\, \, terms}{sum\, of\, n\, \, terms }=\frac{m^2}{n^2}$

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

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

Put $m=2m-1\, \, and\, \, n=2n-1$ , we get

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

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

$\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{a+(m-1)d}{a+(n-1)d}$

From equation (1) ,we get

$\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{2m-1}{2n-1}$

Hence proved.

Question:13 If the sum of n terms of an A.P. is $3 n^2 + 5 n$ 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 $3 n^2 + 5 n$ 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 = $3 n^2 + 5 n$

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

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

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

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,

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

$\Rightarrow \, \9(m+1)+1890=5(m+1)+150m-150$

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

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

$\Rightarrow \, 2044=146m$

$\Rightarrow \, m=14$

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 , $a=100\, \, and\, \, \, d=5$

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

$a_3_0=100+(30-1)5$

$a_3_0=100+(29)5$

$a_3_0=100+145$

$a_3_0=245$

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

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

Answer:

The angles of polygon forms AP with common difference of $5 \degree$ and first term as $120 \degree$ .

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.

Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2:-

NCERT solutions for Class 11 Maths chapter 9 exercise 9.2 are designed by the subject matter experts based on the guideline given by CBSE
As you are familiar with arithmetic progression, it won't take much effort to get command on Class 11 Maths chapter 9 exercise 9.2
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Frequently Asked Questions (FAQs)

1. If a constant term is added to each term of an A.P. the resulting sequence is ?

When a constant term is added to each term of an A.P. the resulting sequence is also an A.P.

2. If a constant term is subtracted from the each term of an A.P. the resulting sequence is ?

When a constant term is subtracted from each term of an A.P. the resulting sequence is also an A.P.

3. Find the 5th term of A.P. 20, 18, 16, ...?

Given a= 20

d = 18 -20 = -2

a_5 = a + 4d

a_5 = 20 -2(4) = 20 - 8 = 12

4. Find the sum first 10 natural numbers ?

n = 10

S_n = n(n+1)/2

S_(10) = 10x11/2 = 550

5. Find the sum of first 5 term of A.P. 20, 18, 16, ...?

Given a = 20

d = 18 -20 = -2

S_n = n/2[2a + (n-1)d]

S_5 = 5/2[40 + 4x(-2)]

S_5 = 5/2[32]

S_5 = 80

6. What is the weightage of Sequences and Series in the CBSE Class 11 Maths ?https://school.careers360.com/ncert/ncert-books

CBSE doesn't provide the chapter-wise marks distributions for CBSE Class 11 Maths exam. The whole unit algebra has 30 marks weightage out of 80 marks in the CBSE 11 Maths exam.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2- Download Free PDF

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Access Sequences And Series Class 11 Chapter 9 Exercise: 9.2

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

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