# NCERT Solutions for Class 11 Maths Chapter 9

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.1, 9.2, 9.3, 9.4 & Miscellaneous are given below to free download in PDF form updated for new academic session 2020-21 free based on latest NCERT Books for 2020-2021.

11th Maths Exercise 9.1, Exercise 9.2, Exercise 9.3, Exercise 9.4 and Miscellaneous Exercise with Supplementary Exercise 9.4 are also available to study online free or session 2020-2021. After passing 10th standard, if someone wants to do directly 12th class, go for NIOS Online Admission. These NCERT Solutions 2020-21 and Offline Apps are appropriate for CBSE as well as MP, UP Board (intermediate) for the academic session 2020-21 onward.## NCERT Solutions for Class 11 Maths Chapter 9

Class: | 11 |

Subject: | Maths |

Chapter 9: | Sequences and Series |

### 11th Maths Chapter 9 Solutions

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series all exercises with supplementary and miscellaneous exercise are given below in PDF form updated for new academic session 2020-2021. Join the discussion forum to ask your doubts and share your knowledge with others.

### 11th Maths Chapter 9 Solutions in English Medium

### 11th Maths Chapter 9 Solutions in PDF

- Download Class 11 Maths Exercise 9.1 Solutions
- Download Class 11 Maths Exercise 9.2 Solutions
- Download Class 11 Maths Exercise 9.3 Solutions
- Download Class 11 Maths Exercise 9.4 Solutions
- Download Class 11 Maths Supplementary Exercise 9.4 Solutions
- Download Class 11 Maths Miscellaneous Exercise 9 Solutions
- Class 11 Maths Solutions Main Page

##### When is a sequence said to be a progression?

A sequence is said to be a progression if the term of the sequence can be expressed by some formula.

##### What is meant by real sequence?

A sequence whose range is a subset of R is called a real sequence.

#### 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.

##### About 11th Maths Chapter 9

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.

### Important Questions on 11th Maths Chapter 9

Substituting n = 1, 2, 3, 4, and 5, we have

a_1=1(1+2)=3

a_2=2(2+2)=8

a_3=3(3+2)=15

a_4=4(4+2)=24

a_5=5(5+2)=35

Hence, the required terms are 3, 8, 15, 24 and 35.

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)[550]

= 98450

Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

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.

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)

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.

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.

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.

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,…….

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.

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.

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.

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