# NCERT Solutions for Class 11 Maths Chapter 9

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (अनुक्रम तथा श्रेणी) Exercise 9.1 or Exercise 9.2 or Exercise 9.3 or Exercise 9.4 or Miscellaneous Exercise with Supplementary Exercise 9.4 are to study online or download free in PDF format. After passing 10th standard, if someone wants to do directly 12th class, go for NIOS Online Admission. These NCERT Solutions and Offline Apps are appropriate for CBSE as well as MP, UP Board (intermediate) for the academic session 2018-19 onward.

 Class: 11 Subject: Maths Chapter 9: Sequences and Series

## NCERT Solutions for Class 11 Maths Chapter 9

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### Class 11 Maths Chapter 9 Sequences and Series Solutions

##### 11 Maths Exercise 9.1 Solutions

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.1 is given below. For other questions, please visit to Exercise 9.2 or Exercise 9.3 or Exercise 9.4 or Miscellaneous Exercise with Supplementary Exercise 9.4 Solutions. Visit to Class 11 Maths main page or move to Top of the page.    ##### 11 Maths Exercise 9.2 Solutions

Class 11 Maths Chapter 9 Sequences and Series Exercise 9.2 is given below. For other questions, please visit to Exercise 9.1 or Exercise 9.3 or Exercise 9.4 or Miscellaneous Exercise with Supplementary Exercise 9.4 Solutions. Visit to Class 11 Maths main page or move to Top of the page.       ##### 11 Maths Exercise 9.3 Solutions

NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3 is given below. For other questions, please visit to Exercise 9.1 or Exercise 9.2 or Exercise 9.4 or Miscellaneous Exercise with Supplementary Exercise 9.4 Solutions. Visit to Class 11 Maths main page or move to Top of the page.          ##### 11 Maths Exercise 9.4 Solutions

11 Maths Chapter 9 Exercise 9.4 Solutions are given below. For other questions, please visit to Exercise 9.1 or Exercise 9.2 or Exercise 9.3 or Miscellaneous Exercise with Supplementary Exercise 9.4 Solutions. Visit to Class 11 Maths main page or move to Top of the page.   ##### 11 Maths Miscellaneous Exercise 9 Sols

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Exercise 9 is given below. For other questions, please visit to Exercise 9.1 or Exercise 9.2 or Exercise 9.3 or Exercise 9.4 or Supplementary Exercise 9.4 Solutions. Visit to Class 11 Maths main page or move to Top of the page.             ##### 11 Maths Supplementary Exercise 9.4 Sols

Class 11 Maths Chapter 9 Supplementary Exercise 9.4 is given below. For other questions, please visit to Exercise 9.1 or Exercise 9.2 or Exercise 9.3 or Exercise 9.4 or Miscellaneous Exercise Solutions. Visit to Class 11 Maths main page or move to Top of the page.    Visit to Class 11 Maths main page or Top of the page.

###### Important Terms Related to Sequences & Series
• A sequence is a function whose domain is the set N of natural numbers or some subset of it.
• In an A.P., the sum of the terms equidistant from the beginning and from the end is always same, and equal to the sum of the first and the last term.
• If three terms of A.P. are to be taken then we choose then as a – d, a, a + d.
• If four terms of A.P. are to be taken then we choose then as a – 3d, a – d, a + d, a + 3d.
• If five terms of A.P are to be taken, then we choose then as: a – 2d, a – d, a, a + d, a + 2d.

• A sequence is said to be a progression if the term of the sequence can be expressed by some formula.
• A sequence whose range is a subset of R is called a real sequence.
• In a G.P., the product of the terms equidistant from the beginning and from the end is always same and equal to the product of the first and the last term.
• If each term of a G.P. be raised to some power then the resulting terms are also in G.P.
• If a, b, c are in A.P. then 2b = a + c.
• If a, b, c are in G.P. then b² = ac.

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

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, 115 … 995.
Here, first term, a = 105 and common difference, d = 5
Now,a_n=a+(n-1)d
⇒ 995 = 105+(n-1)×5
⇒ 890=(n-1)×5
⇒ 178=(n-1)
⇒ n=179
S_n = n/2 [2a+(n-1)d] ⇒ S_179 = 179/2 [2×105+(179-1)×5] ⇒ S_179 = (179) = 98450
Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

#### 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.

First term = 2 and let d be the common difference of the A.P.
Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d …
Sum of first five terms = 10 + 10d
Sum of next five terms = 10 + 35d
According to the given condition,
10+10d=1/4 (10+35d)
⇒ 40+40d=10+35d
⇒ 5d=-30
⇒ d=-6
Now,a_20=a+(20-1)d
⇒ a_20=2+19×(-6)=-112
Hence, the 20th term of the A.P. is –112.

#### Find the sum to n terms of the A.P., whose k^th term is 5k+1.

It is given that the kth term of the A.P. is 5k + 1.
k^th term = a_k= a + (k – 1)d
∴ a + (k – 1)d = 5k + 1
a + kd – d = 5k + 1
Comparing the coefficient of k, we obtain d = 5
Therefore, a – d = 1
⇒ a – 5 = 1
⇒ a = 6
S_n = n/2 [2a+(n-1)d] = n/2 [2×6+(n-1)×5] = n/2 (5n+7)

#### If the sum of n terms of an A.P. is (pn + qn^2), where p and q are constants, find the common difference.

We know that: S_n=n/2 [2a+(n-1)d] According to question,
n/2 [2a + (n-1)d] = pn + qn^2
⇒ na + n^2/2 d – n/2 d = pn + qn^2
Comparing the coefficients of n2 on both sides, we have
d/2 = q
⇒ d = 2q
Hence, the common difference of the A.P. is 2q.

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

Let A1, A2, A3, A4 and A5 be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 is an A.P.
Here, a = 8, b = 26, n = 7
Therefore, 26 = 8 + (7 – 1) d
⇒ 6d = 26 – 8 = 18
⇒ d = 3
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14,
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20,
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23
Hence, the required five numbers between 8 and 26 are 11, 14, 17, 20 and 23.

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

The first installment of the loan is ₹ 100. The second installment of the loan is ₹ 105 and so on.
The amount that the man repays every month forms an A.P.
The A.P. is 100, 105, 110 …
First term, a = 100
Common difference, d = 5
A30 = a + (30 – 1)d
= 100 + (29) (5)
= 100 + 145 = 245
Hence, the amount to be paid in the 30th installment is ₹ 245.

#### Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Let the first term of the given GP is a and the common ratio is r.
Sum of the first two terms is – 4,
therefore, a_1+a_2=-4
⇒ a + ar = -4 … (1)
The fifth term is 4 times the third term,
therefore, a_5=4a_3
⇒ ar^4 = 4×ar^2
⇒ r^2=4
⇒ r=±2
If r=2,from (1),we have a + a×2 =-4
⇒ 3a = -4
⇒ a =-4/3
Therefore,the required GP:
a,ar,ar^2,….is given by
-4/3,-4/3×2,-4/3×2^2,….
or-4/3,-8/3,-16/3,…….
If r=-2,from (1),
we have a+a×(-2)=-4
⇒-a=-4
⇒a=4
Therefore,the required GP: a,ar,ar^2,….is given by 4,4×(-2),4×(-2)^2,….or 4,-8,16,…….

#### Insert two number between 3 and 81 so that the resulting sequence is G.P.

Let G1 and G2 be two numbers between 3 and 81 such that the series, 3, G1, G2, 81 forms a G.P.
Let a be the first term and r be the common ratio of the G.P.
∴ 81 = (3) (r)^3
⇒ r^3=27
∴ r = 3 (Taking real roots only)
For r = 3,
G_1= ar = (3) (3) = 9 and G_2= ar^2= (3) (3)^2 = 27
Thus, the required two numbers are 9 and 27.

#### If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Let the root of the quadratic equation be a and b.
According to the question,
AM=(a+b)/2=8
⇒a+b=16 … (1)
GM=√ab=5
⇒ab=25 … (2)
The quadratic equation is given by,
x^2– x(Sum of roots) + (Product of roots) = 0
⇒x^2 – x (a + b) + (ab) = 0
⇒x^2 – 16x + 25 = 0
[Using (1) and (2)] Hence, the required quadratic equation is x^2 – 16x + 25 = 0.

#### If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Let the three numbers in A.P. be a – d, a, and a + d.
According to question,
(a – d)+ (a)+ (a + d)= 24 … (1)
⇒ 3a = 24 ⇒ a = 8
Now, product of the numbers:
(a – d)a (a + d)= 440 … (2)
⇒ (8 – d)(8)(8 + d)= 440
⇒ (8 – d)(8 + d)= 55
⇒ 64 – d2 = 55
⇒ d2 = 64 – 55 = 9
⇒ d = ± 3
Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.
Hence, the three numbers are 5, 8 and 11.

#### 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Let x be the number of days in which 150 workers finish the work.
According to the given information,
150x = 150 + 146 + 142 + ….(x + 8) terms
The series 150 + 146 + 142 + …. (x + 8) terms is an A.P. with first term 146, common difference –4 and number of terms as (x + 8).
⇒150x=(x+8)/2 [2×150+(x+8-1)×(-4)] ⇒300x=(x+8)(300-4x-28)
⇒300x=(x+8)(272-4x)
⇒300x=272x-4x^2+2176-32x
⇒4x^2+60x-2176=0
⇒x^2+15x-544=0
⇒x^2+32x-17x-544=0
⇒x(x+32)-17(x+32)=0
⇒(x+32)(x-17)=0
⇒x=-32 or 17
However, x cannot be negative.
∴ x = 17
Therefore, originally, the number of days in which the work was completed is 17.
Hence, required number of days = (17 + 8) = 25

## 2 thoughts on “NCERT Solutions for Class 11 Maths Chapter 9”

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