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

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

Edited By Vishal kumar | Updated on Nov 08, 2023 08:06 AM IST

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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  1. NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2- Download Free PDF
  2. Download the PDF of NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2
  3. Access Sequences And Series Class 11 Chapter 9 Exercise: 9.2
  4. Question:1 Find the sum of odd integers from 1 to 2001.
  5. More About NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2:-
  6. Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2:-
  7. Key Features of NCERT 11th Class Maths Exercise 9.2 Answers
  8. NCERT Solutions of Class 11 Subject Wise
  9. Subject Wise NCERT Exampler Solutions
NCERT Solutions for Exercise 9.2 Class 11 Maths Chapter 9 - Sequences and Series
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.

Also, see

**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.

More About NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2:-

Class 11 Maths chapter 9 exercise 9.2 consists of questions related to finding the general terms of arithmetic progression, the sum of the n terms of arithmetic progression, arithmetic mean of the progression, etc. There are some properties of arithmetic progression like adding and subtracting the constant from each term of the progression, multiplying and dividing by the non zero constant to each term of the progression, etc given before the Class 11 Maths chapter 9 exercise 9.2. There are five solved examples given before this exercise. You must go through these properties of arithmetic progression and solved examples before solving exercise 9.2 Class 11 Maths.

Also Read| Sequences And Series Class 11 Notes

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
  • You can use Class 11 Maths chapter 9 exercise 9.2 solutions as revision notes to quickly revise the important concepts.
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Key Features of NCERT 11th Class Maths Exercise 9.2 Answers

  1. Comprehensive Scope: These ex 9.2 class 11 answers address all the exercises 9.2 problems of the NCERT 11th Class Mathematics textbook.

  2. Step-by-Step Approach: The solutions are presented in a step-by-step manner, making it easier for students to understand and follow the problem-solving process.

  3. Clarity and Precision: The class 11 ex 9.2 answers are written with utmost clarity and precision, ensuring that students can grasp the mathematical concepts and techniques required for effective problem-solving.

  4. Appropriate Mathematical Notation: These 11th class maths exercise 9.2 answers use the correct mathematical notations and terminology, aiding students in becoming proficient in the language of mathematics.

  5. Free Accessibility: Typically, these class 11 maths chapter 9 exercise 9.2 solution are available free of charge, allowing students to access them without any cost, providing a valuable resource for self-study.

  6. Supplementary Learning Aid: These ex 9.2 class 11 answers can serve as a supplementary learning resource to reinforce classroom teachings and assist students in preparing for exams.

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

Also see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Question (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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Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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
Read More

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