NCERT Solution Class 11 Chapter 9- Sequences and Series

Practising the NCERT Solutions Class 11 Chapter 9 Sequences and Series can help the students develop a thorough understanding of the topics explained in the Chapter. These solutions prepared by highly experienced teachers at BYJU’S are bound to help the students in addressing challenging questions with utmost confidence. Using the NCERT Solutions provided here, students can learn new methods of solving a particular problem in expeditious time to improve their performance in the Maths exam. These solutions have been designed after undertaking extensive research on each question and their problem solving method.

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Exercises :
Exercise 9.1 Solutions 14 Questions

Exercise 9.2 Solutions 18 Questions

Exercise 9.3 Solutions 32 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise On Chapter 9 Solutions 32 Questions

Access Answers of Maths NCERT Class 11 Chapter 9 – Sequences and Series

Exercise 9.1 Page No: 180

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. an = n (n + 2) 

Solution:

Given,

nth term of a sequence an = n (n + 2) 

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a1= 1(1 + 2) = 3

a2= 2(2 + 2) = 8

a3= 3(3 + 2) = 15

a4= 4(4 + 2) = 24

a5= 5(5 + 2) = 35

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

2. an = n/n+1

Solution:

Given nth term, an = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 1

Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.

3. an = 2n

Solution:

Given nth term, an = 2n

On substituting n = 1, 2, 3, 4, 5, we get

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

Hence, the required terms are 2, 4, 8, 16, and 32.

4.  an = (2n – 3)/6

Solution:

Given nth term, an = (2n – 3)/6

On substituting = 1, 2, 3, 4, 5, we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 2

Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..

5. an = (-1)n-1 5n+1

Solution:

Given nth term, an = (-1)n-1 5n+1

On substituting = 1, 2, 3, 4, 5, we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 3

Hence, the required terms are 25, –125, 625, –3125, and 15625.

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 4

6.

Solution:

On substituting n = 1, 2, 3, 4, 5, we get first 5 terms

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 5

Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

7. an = 4n – 3; a17, a24

Solution:

Given,

nth term of the sequence is an = 4n – 3

On substituting n = 17, we get

a17 = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a24 = 4(24) – 3 = 96 – 3 = 93

8. an = n2/2n ; a7

Solution:

Given,

nth term of the sequence is an = n2/2n

Now, on substituting n = 7, we get

a7 = 72/27 = 49/ 128

9. an = (-1)n-1 n3; a9

Solution:

Given,

nth term of the sequence is an = (-1)n-1 n3

On substituting n = 9, we get

a9 = (-1)9-1 (9)3 = 1 x 729 = 729

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 6

10.

Solution:

On substituting n = 20, we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 7

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

11. a1 = 3, an = 3an-1 + 2 for all n > 1

Solution:

Given, an = 3an-1 + 2 and a1 = 3

Then,

a2 = 3a1 + 2 = 3(3) + 2 = 11

a3 = 3a2 + 2 = 3(11) + 2 = 35

a4 = 3a3 + 2 = 3(35) + 2 = 107

a5 = 3a4 + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

12. a1 = -1, an = an-1/n, n ≥ 2

Solution:

Given,

an = an-1/n and a1 = -1

Then,

a2 = a1/2 = -1/2

a3 = a2/3 = -1/6

a4 = a3/4 = -1/24

a5 = a4/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

13. a1 = a2 = 2, an = an-1 – 1, n > 2

Solution:

Given,

a1 = a2, an = an-1 – 1

Then,

a3 = a2 – 1 = 2 – 1 = 1

a4 = a3 – 1 = 1 – 1 = 0

a5 = a4 – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

14. The Fibonacci sequence is defined by

1 = a1 = a2 and an = an – 1 + an – 2, n > 2

Find an+1/an, for n = 1, 2, 3, 4, 5 

Solution:

Given,

1 = a1 = a2

an = an – 1 + an – 2, n > 2

So,

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

Thus,

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 8


Exercise 9.2 Page No: 185

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

Solution:

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, a = 1

Common difference, d = 2

Now,

a + (n -1)d = 2001

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

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 1

Therefore, the sum of odd numbers from 1 to 2001 is 1002001.

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

Solution:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where, the first term, a = 105

Common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 2

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

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.

Solution:

Given,

The first term (a) of an A.P = 2

Let’s assume d be the common difference of the A.P.

So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20th term of the A.P. is –112.

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

Solution:

Let’s consider the sum of n terms of the given A.P. as –25.

We known that,

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

where n = number of terms, a = first term, and d = common difference

So here, a = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 3

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 ½ (pq + 1) where p ≠ q.  

Solution:

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 4

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 5

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

Solution:

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

Sn = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n2

3n2 – 53n + 232 = 0

3n2 – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, 8th term is the last term of the A.P.

a8 = 25 + (8 – 1)(-3)

= 25 – 21

= 4

7. Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Solution:

Given, the kth term of the A.P. is 5k + 1.

kth term = ak = + (k – 1)d

And,

+ (k – 1)d = 5k + 1

a + kd – d = 5k + 1

On comparing the coefficient of k, we get d = 5

– = 1

a – 5 = 1

⇒ a = 6

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 6

8. If the sum of n terms of an A.P. is (pn qn2), where p and q are constants, find the common difference.

Solution:

We know that,

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

From the question we have,

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 7

On comparing the coefficients of n2 on both sides, we get

d/2 = q

Hence, d = 2q

Therefore, the common difference of the A.P. is 2q.

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.

Solution:

Let a1a2, and d1dbe the first terms and the common difference of the first and second arithmetic progression respectively.

Then, from the question we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 8

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.

Solution:

Let’s take a and d to be the first term and the common difference of the A.P. respectively.

Then, it given that

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 9

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 10

Therefore, the sum of (p + q) terms of the A.P. is 0.

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 11Prove that 

Solution:

Let a1 and d be the first term and the common difference of the A.P. respectively.

Then according to the question, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 12

Now, subtracting (2) from (1), we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 13

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 14

12. The ratio of the sums of m and n terms of an A.P. is m2n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).

Solution:

Let’s consider that a and b to be the first term and the common difference of the A.P. respectively.

Then from the question, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 15

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 16

Hence, the given result is proved.

13. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

Solution:

Let’s consider a and b to be the first term and the common difference of the A.P. respectively.

am = a + (m – 1)d = 164 … (1)

We the sum of the terms is given by,

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 17

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

Solution:

Let’s assume A1, A2, A3, A4, and A5 to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.

Here we have, 

= 8, = 26, n = 7

So,

26 = 8 + (7 – 1) d

6d = 26 – 8 = 18

= 3

Now,

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 + 4= 8 + 4 × 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

15. If NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 18is the A.M. between a and b, then find the value of n.

Solution:

The A.M between a and b is given by, (a + b)/2

Then according to the question,

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 19

Thus, the value of n is 1.

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 7th and (m – 1)th numbers is 5: 9. Find the value of m.

Solution:

Let’s consider a1, a2, … am be m numbers such that 1, a1, a2, … am, 31 is an A.P.

And here, 

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = 30/ (m + 1) ……. (1)

Now,

a1 = a + d

a2 = a + 2d

a3 = a + 3d …

Hence, a7 = a + 7d

am–1 = a + (m – 1) d

According to the question, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 20

Therefore, the value of m is 14.

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?

Solution:

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105 and so on as the instalment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And the, A.P. is 100, 105, 110, …

Where, first term, a = 100

Common difference, d = 5

So, the 30th term in this A.P. will be

A30  = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30th instalment will be Rs 245.

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

Solution:

It’s understood from the question that, the angles of the polygon will form an A.P. with common difference d = 5° and first term a = 120°.

And, we know that the sum of all angles of a polygon with n sides is 180° (n – 2).

Thus, we can say

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.2 - 21

Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.


Exercise 9.3 Page No: 192

1. Find the 20th and nthterms of the G.P. 5/2, 5/4, 5/8, ………

Solution:

Given G.P. is 5/2, 5/4, 5/8, ………

Here, a = First term = 5/2

r = Common ratio = (5/4)/(5/2) = ½

Thus, the 20th term and nth term

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 1

2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Solution:

Given,

The common ratio of the G.P., r = 2

And, let a be the first term of the G.P.

Now,

a8 = ar 8–1 = ar7

ar7 = 192

a(2)7 = 192

a(2)7 = (2)6 (3)

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 2

3. The 5th, 8th and 11th terms of a G.P. are pq and s, respectively. Show that q2 = ps.

Solution:

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then according to the question, we have

a5 = a r5–1 a r4 = p … (i)

aa r8–1 a r7 = q … (ii)

a11 = a r11–1 a r10 = … (iii)

Dividing equation (ii) by (i), we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 3

4. The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.

Solution:

Let’s consider a to be the first term and r to be the common ratio of the G.P.

Given, a = –3

And we know that, 

an = arn–1

So, aar3 = (–3) r3

a2 = a r1 = (–3) r

Then from the question, we have

(–3) r3 = [(–3) r]2

⇒ –3r3 = 9 r2

⇒ r = –3

a7 = a r 7–1 a r6 = (–3) (–3)6 = – (3)7 = –2187

Therefore, the seventh term of the G.P. is –2187.

5. Which term of the following sequences:

(a) 2, 2√2, 4,… is 128 ? (b) √3, 3, 3√3,… is 729 ?

(c) 1/3, 1/9, 1/27, … is 1/19683 ?

Solution:

(a) The given sequence, 2, 2√2, 4,…

We have,

a = 2 and r = 2√2/2 = √2

Taking the nth term of this sequence as 128, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 4

Therefore, the 13th term of the given sequence is 128.

(ii) Given sequence, √3, 3, 3√3,…

We have,

a = √3 and r = 3/√3 = √3

Taking the nth term of this sequence to be 729, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 5

Therefore, the 12th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, …

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the nth term of this sequence to be 1/19683, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 6

Therefore, the 9th term of the given sequence is 1/19683.

6. For what values of x, the numbers -2/7, x, -7/2 are in G.P?

Solution:

The given numbers are -2/7, x, -7/2.

Common ratio = x/(-2/7) = -7x/2 

Also, common ratio = (-7/2)/x = -7/2x

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 7

Therefore, for x = ± 1, the given numbers will be in G.P.

7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

Solution:

Given G.P., 0.15, 0.015, 0.00015, …

Here, a = 0.15 and r = 0.015/0.15 = 0.1 

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 8

8. Find the sum to n terms in the geometric progression √7, √21, 3√7, ….

Solution:

The given G.P is √7, √21, 3√7, ….

Here,

a = √7 and

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 9

9. Find the sum to n terms in the geometric progression 1, -a, a2, -a3 …. (if a ≠ -1)

Solution:

The given G.P. is 1, -a, a2, -a3 ….

Here, the first term = a1 = 1

And the common ratio = r = – a

We know that,

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 10

10. Find the sum to n terms in the geometric progression x3, x5, x7, … (if x ≠ ±1 )

Solution:

Given G.P. is x3, x5, x7, …

Here, we have a = x3 and r = x5/x3 = x2

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 11

11. Evaluate: NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 12

Solution:

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 13

12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Solution:

Let a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 …… (1)

(a/r) (a) (ar) = 1 …….. (2)

From (2), we have

a3 = 1

Hence, a = 1 [Considering real roots only]

Substituting the value of a in (1), we get

1/r + 1 + r = 39/10

(1 + r + r2)/r = 39/10

10 + 10r + 10r2 = 39r

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(5r – 2) (2r – 5) = 0

Thus,

r = 2/5 or 5/2

Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Solution:

Given G.P. is 3, 32, 33, …

Let’s consider that n terms of this G.P. be required to obtain the sum of 120.

We know that,

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 14

Here, a = 3 and r = 3

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 15

Equating the exponents we get, n = 4

Therefore, four terms of the given G.P. are required to obtain the sum as 120.

14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Solution:

Let’s assume the G.P. to be aarar2ar3, …

Then according to the question, we have

ar + ar2 = 16 and arar4 + ar= 128

a (1 + r + r2) = 16 … (1) and,

ar3(1 + r + r2) = 128 … (2)

Dividing equation (2) by (1), we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 16

r3 = 8

r = 2

Now, using r = 2 in (1), we get

a (1 + 2 + 4) = 16

a (7) = 16

a = 16/7

Now, the sum of terms is given as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 17

15. Given a G.P. with a = 729 and 7th term 64, determine S7.

Solution:

Given,

a = 729 and a7 = 64

Let r be the common ratio of the G.P.

Then we know that, an = a rn–1

a7 = ar7–1 = (729)r6

⇒ 64 = 729 r6

r6 = 64/729

r6 = (2/3)6

r = 2/3

And, we know that

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 18

= 37 – 27

= 2187 – 128

= 2059

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

Solution:

Consider a to be the first term and r to be the common ratio of the G.P.

Given, S2 = -4

Then, from the question we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 19

And,

a5 = 4 x a3

ar4 = 4ar2

r2 = 4

r = ± 2

Using the value of r in (1), we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 20

Therefore, the required G.P is

-4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……

17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Solution:

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a4 = a r3 = x … (1)

a10 = a r9 = y … (2)

a16 = a r15 z … (3)

On dividing (2) by (1), we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 21

18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Solution:

Given sequence: 8, 88, 888, 8888…

This sequence is not a G.P.

But, it can be changed to G.P. by writing the terms as

Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 22

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

Solution:

The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½

= 64[4 + 2 + 1 + ½ + 1/22]

Now, it’s seen that

4, 2, 1, ½, 1/22 is a G.P.

With first term, a = 4

Common ratio, r =1/2

We know,

 
NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 42

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 23

Therefore, the required sum = 64(31/4) = (16)(31) = 496

20. Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn-1 and A, AR, AR2, … ARn-1 form a G.P, and find the common ratio.

Solution:

To be proved: The sequence, aAarARar2AR2, …arn–1ARn–1, forms a G.P.

Now, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 24

Therefore, the above sequence forms a G.P. and the common ratio is rR.

21. Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Solution:

Consider a to be the first term and r to be the common ratio of the G.P.

Then,

a1 = aa2 = ara3 = ar2a4 = ar3

From the question, we have

a3 = a1 + 9

ar2 = a + 9 … (i)

a2 = a4 + 18

ar ar3 + 18 … (ii)

So, from (1) and (2), we get

a(r2 – 1) = 9 … (iii)

ar (1– r2) = 18 … (iv)

Now, dividing (4) by (3), we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 25

-r = 2

r = -2

On substituting the value of r in (i), we get

4a + 9

3a = 9

∴ a = 3

Therefore, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3 

i.e., 3¸–6, 12, and –24.

22. If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-r br-p cp-q = 1

Solution:

Let’s take A to be the first term and R to be the common ratio of the G.P.

Then according to the question, we have

ARp–1 a

ARq–1 b

ARr–1 c

Then,

aq–r br–p cp–q

Aq× R(p–1) (q–r) × Arp × R(q–1) (rp) × Apq × R(–1)(pq)

Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r p – pq) + (pr – p – qr + q)

A0 × R0

= 1

Hence proved.

23. If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.

Solution:

Given, the first term of the G.P is a and the last term is b.

Thus,

The G.P. is aarar2ar3, … arn–1, where r is the common ratio.

Then,

b = arn–1  … (1)

P = Product of n terms

= (a) (ar) (ar2) … (arn–1)

= (a × a ×…a) (r × r2 × …rn–1)

an r 1 + 2 +…(n–1)  … (2)

Here, 1, 2, …(n – 1) is an A.P.

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 26

And, the product of n terms P is given by,

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 27

24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 28.

Solution:

Let a be the first term and be the common ratio of the G.P.

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 30

Since there are n terms from (n +1)th to (2n)th term,

Sum of terms from(n + 1)th to (2n)th term 
NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 31

a +1 = ar n + 1 – 1 = arn

Thus, required ratio = 
NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 32
NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 43

Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 33.

25. If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.

Solution:

Given, abcd are in G.P.

So, we have

bc = ad  … (1)

b2 = ac  … (2)

c2 = bd  … (3)

Taking the R.H.S. we have

R.H.S.

= (ab + bc + cd)2

= (ab + ad cd)2  [Using (1)]

= [ab + d (a + c)]2

a2b2 + 2abd (a + c) + d2 (a + c)2

a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)

a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2  [Using (1) and (2)]

a2b2 + a2c2 + a2c2 + b2cb2c2 + d2a2 + d2b2 + d2b2 + d2c2

a2b2 + a2c2 + a2db× b2 + b2c2 + b2d2 + c2b2 + c× c2 + c2d2

[Using (2) and (3) and rearranging terms]

a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

(a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2

26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Solution:

Let’s assume G1 and G2 to be two numbers between 3 and 81 such that the series 3, G1G2, 81 forms a G.P.

And let a be the first term and r be the common ratio of the G.P.

Now, we have the 1st term as 3 and the 4th term as 81.

81 = (3) (r)3

r3 = 27

∴ r = 3 (Taking real roots only)

For r = 3,

G1 = ar = (3) (3) = 9

G2 = ar2 = (3) (3)2 = 27

Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.

27. Find the value of n so that NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 34may be the geometric mean between a and b.

Solution:

We know that,

The G. M. of a and b is given by √ab.

Then from the question, we have

 
NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 35

By squaring both sides, we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 36

28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratioNCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 37.

Solution:

Consider the two numbers be a and b.

Then, G.M. = √ab.

From the question, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 38

29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 39

numbers are .

Solution:

Given that A and G are A.M. and G.M. between two positive numbers.

And, let these two positive numbers be a and b.

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 40

30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

Solution:

Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.

Here we have, a = 30 and r = 2

So, a3 = ar2 = (30) (2)2 = 120

Thus, the number of bacteria at the end of 2nd hour will be 120.

And, a5 = ar4 = (30) (2)4 = 480

The number of bacteria at the end of 4th hour will be 480.

an +1 arn = (30) 2n

Therefore, the number of bacteria at the end of nth hour will be 30(2)n.

31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Solution:

Given,

The amount deposited in the bank is Rs 500.

At the end of first year, amount = Rs 500(1 + 1/10) = Rs 500 (1.1)

At the end of 2nd year, amount = Rs 500 (1.1) (1.1)

At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on….

Therefore,

The amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)10

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

Solution:

Let’s consider the roots of the quadratic equation to be a and b.

Then, we have

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.3 - 41

We know that,

A quadratic equation can be formed as,

x2 – x (Sum of roots) + (Product of roots) = 0

x2 – x (a + b) + (ab) = 0

x2 – 16x + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation is x2 – 16x + 25 = 0


Exercise 9.4 Page No: 196

Find the sum to n terms of each of the series in Exercises 1 to 7.

1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Solution:

Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

It’s seen that,

nth term, an = n ( n + 1)

Then, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 1

2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Solution:

Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

It’s seen that,

nth term, an = n ( n + 1) ( n + 2)

= (n2 + n) (n + 2)

= n+ 3n+ 2n

Then, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 2

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 3

3. 3 × 12 + 5 × 22 + 7 × 32 + …

Solution:

Given series is 3 ×12 + 5 × 22 + 7 × 32 + …

It’s seen that,

nth term, an = ( 2n + 1) n2 = 2n3 + n2

Then, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 4

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 5

4. Find the sum to n terms of the series 

Solution:

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 6

5. Find the sum to n terms of the series 52 + 62 + 72 + … + 202

Solution:

Given series is 52 + 62 + 72 + … + 202

It’s seen that,

nth term, an = ( n + 4)2 = n2 + 8n + 16

Then, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 7

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 8

6. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Solution:

Given series is 3 × 8 + 6 × 11 + 9 × 14 + …

It’s found out that,

a= (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n2 + 15n

Then, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 9

7. Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …

Solution:

Given series is 12 + (12 + 22) + (12 + 2+ 32 ) + …

Finding the nth term, we have

an = (12 + 22 + 32 +…….+ n2)


NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 10

Now, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 11

8. Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).

Solution:

Given,

an = n (n + 1) (n + 4) = n(n+ 5n + 4) = n3 + 5n2 + 4n

Now, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 12

9. Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Solution:

Given,

nth term of the series as:

an = n2 + 2n

Then, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 13

10. Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Solution:

Given,

nth term of the series as:

an = (2n – 1)2 = 4n2 – 4n + 1

Then, the sum of n terms of the series can be expressed as

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.4 - 14


Miscellaneous Exercise Page No: 199

1. Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Solution:

Let’s take a and d to be the first term and the common difference of the A.P. respectively.

We know that, the kth term of an A. P. is given by

ak = a + (k –1) d

So, am + n = a + (m + n –1) d

And, am – n = a + (m – n –1) d

am a + (m –1) d

Thus,

am + n + am – n  = a + (m + n –1) d + a + (m – n –1) d

= 2a + (m + n –1 + m – n –1) d

= 2a + (2m – 2) d

= 2a + 2 (m – 1) d

=2 [a + (m – 1) d]

= 2am

Therefore, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term

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

Solution:

Let’s consider the three numbers in A.P. as a – da, and a + d.

Then, from the question we have

(a – d) + (a) + (a + d) = 24 … (i)

3a = 24

∴ a = 8

And,

(a – da (a + d) = 440 … (ii)

(8 – d) (8) (8 + d) = 440

(8 – d) (8 + d) = 55

64 – d2 = 55

d2 = 64 – 55 = 9

= ± 3

Thus,

When d = 3, the numbers are 5, 8, and 11 and

When d = –3, the numbers are 11, 8, and 5.

Therefore, the three numbers are 5, 8, and 11.

3. Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)

Solution:

Let’s take a and d to be the first term and the common difference of the A.P. respectively.

So, we have

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 1

4. Find the sum of all numbers between 200 and 400 which are divisible by 7.

Solution:

First let’s find the numbers between 200 and 400 which are divisible by 7.

The numbers are:

203, 210, 217, … 399

Here, the first term, a = 203

Last term, l = 399 and

Common difference, d = 7

Let’s consider the number of terms of the A.P. to be n.

Hence, an = 399 = a + (n –1) d

399 = 203 + (n –1) 7

7 (n –1) = 196

n –1 = 28

n = 29

Then, the sum of 29 terms of the A.P is given by:

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 2

Therefore, the required sum is 8729.

5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Solution:

First let’s find the integers from 1 to 100, which are divisible by 2.

And, they are 2, 4, 6… 100.

Clearly, this forms an A.P. with the first term and common difference both equal to 2.

So, we have

100 = 2 + (n –1) 2

n = 50

Hence, the sum is

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 3

Now, the integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This also forms an A.P. with the first term and common difference both equal to 5.

So, we have

100 = 5 + (n –1) 5

5n = 100

n = 20

Hence, the sum is

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 4

Lastly, the integers which are divisible by both 2 and 5, are 10, 20, … 100.

And this also forms an A.P. with the first term and common difference both equal to 10.

So, we have

100 = 10 + (n –1) (10)

100 = 10n

n = 10

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 5

Thus, the required sum = 2550 + 1050 – 550 = 3050

Therefore, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

6. Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Solution:

We have to first find the two-digit numbers, which when divided by 4, yield 1 as remainder.

They are: 13, 17, … 97.

As it’s seen that this series forms an A.P. with first term (a) 13 and common difference (d) 4.

Let n be the number of terms of the A.P.

We know that, the nth term of an A.P. is given by, 

an = a + (n –1) d

So, 97 = 13 + (n –1) (4)

4 (n –1) = 84

n – 1 = 21

n = 22

Now, the sum of n terms of an A.P. is given by,

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 6

Therefore, the required sum is 1210.

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 7

7. If is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that , find the value of n.

Solution:

Given that,

(x + y) = (x) × (y) for all xy ∈ N … (1)

(1) = 3

Taking x = y = 1 in (1), we have

f (1 + 1) = (2) = (1) (1) = 3 × 3 = 9

Similarly,

(1 + 1 + 1) = (3) = (1 + 2) = (1) (2) = 3 × 9 = 27

And, (4) = (1 + 3) = f (1) (3) = 3 × 27 = 81

Thus, (1), (2), (3), …, that is 3, 9, 27, …, forms a G.P. with the first term and common ratio both equal to 3.

We know that sum of terms in G.P is given by,

 
NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 8

And it’s given that,

 
NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 9

Hence, the sum of terms of the function is 120.

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 10

Therefore, the value of n is 4.

8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Solution:

Given that the sum of some terms in a G.P is 315.

Let the number of terms be n.

We know that, sum of terms is

 
NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 11

Given that the first term a is 5 and common ratio is 2.

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 12

Hence, the last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160

Therefore, the last term of the G.P. is 160.

9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Solution:

Let’s consider a and r to be the first term and the common ratio of the G.P. respectively.

Given, a = 1

a3 = ar2 = r2

a5 = ar4 = r4

Then, from the question we have

r2 + r4 = 90

r4 + r2 – 90 = 0

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 13

Therefore, the common ratio of the G.P. is ±3.

10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Solution:

Let’s consider the three numbers in G.P. to be as aar, and ar2.

Then from the question, we have

a + ar + ar2 = 56

a (1 + r + r2) = 56

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 14 … (1)

Also, given

a – 1, ar – 7, ar2 – 21 forms an A.P.

So, (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)

ar – a – 6 = ar– ar – 14

ar– 2ar + a = 8

ar– ar – ar + a = 8

a(r+ 1 – 2r) = 8

(r – 1)2 = 8 … (2)

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 15

7(r2 – 2r + 1) = 1 + r + r2

7r2 – 14 r + 7 – 1 – r – r2 = 0

6r2 – 15r + 6 = 0

6r2 – 12r – 3r + 6 = 0

6r (r – 2) – 3 (r – 2) = 0

(6r – 3) (r – 2) = 0

r = 2, 1/2

When r = 2, a = 8

When r = ½, a = 32 

Thus,

When r = 2, the three numbers in G.P. are 8, 16, and 32.

When r = 1/2, the three numbers in G.P. are 32, 16, and 8.

Therefore in either case, the required three numbers are 8, 16, and 32.

11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Solution:

Let’s consider the terms in the G.P.to be T1, T2, T3, T4, … T2n.

The number of terms = 2n

Then, from the question we have

T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1]

T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0

T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1] …… (1)

Now, let the terms in G.P. be aarar2ar3, …

Then (1) becomes,

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 16[Using sum of terms in G.P.]

ar = 4a

r = 4

Thus, the common ratio of the G.P. is 4.

12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Solution:

Let’s consider the terms in A.P. to be aa + da + 2da + 3d, … a + (n – 2) da + (n – 1)d.

From the question, we have

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]

= 4a + (4n – 10) d

Then according to the given condition,

4a + 6d = 56

4(11) + 6d = 56 [Since a = 11 (given)]

6d = 12

d = 2

Hence, 4a + (4n –10) d = 112

4(11) + (4n – 10)2 = 112

(4n – 10)2 = 68

4n – 10 = 34

4n = 44

n = 11

Therefore, the number of terms of the A.P. is 11.

13. IfNCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 17, then show that abc and d are in G.P.

Solution:

Given,

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 18

On cross multiplying, we have

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 19

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 20Also, given

On cross multiplying, we have

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 21

From (1) and (2), we get

b/a = c/b = d/c

Therefore, a, b, c and d are in G.P.

14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn

Solution:

Let the terms in G.P. be aarar2ar3, … arn – 1

Form the question, we have

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 22
NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 23

Hence, P2 Rn = Sn

15. The pthqth and rth terms of an A.P. are a, b, c respectively.

Show that (q – r) a + (r – p) b + (p -q) c = 0

Solution:

Let’s assume t and d to be the first term and the common difference of the A.P. respectively.

Then the nth term of the A.P. is given by, an + (n – 1d

Thus,

ap = t + (p – 1d = a  … (1)

aq = t + (q – 1) d = b  … (2)

ar = t + (r – 1d = c  … (3)

On subtracting equation (2) from (1), we get

(p – 1 – q + 1) d = a – b

(p – qd = a – b

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 24

On subtracting equation (3) from (2), we get

(q – 1 – r + 1) d = b – c

(q – rd = b – c

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 25

Equating both the values of d obtained in (4) and (5), we get

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 26

-a (q – r) – b (r – p) – c (p – q) = 0

a (q – r) + b (r – p) + c (p – q) = 0

Therefore, the given result is proved.

16. If aNCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 27are in A.P., prove that a, b, c are in A.P.

Solution:

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 28

17. If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.

Solution:

Given, a, b, c,and d are in G.P.

So, we have

b2 = ac … (i)

c2 = bd … (ii)

ad = bc … (iii)

Required to prove (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,

(bn + cn)2 = (an + bn) (cn + dn)

Taking L.H.S.

(bn + cn)2 = b2+ 2bncn + c2n

= (b2)n+ 2bncn + (c2) n

= (ac)n + 2bncn + (bd)n  [Using (i) and (ii)]

an cn + bncnbn cn + bn dn

an cn + bncnan dn + bn dn  [Using (iii)]

cn (an + bn) + dn (an + bn)

= (an + bn) (cn + dn)

= R.H.S.

Therefore, (an + bn), (bn + cn), and (cn + dn) are in G.P

– Hence proved.

18. If a and are the roots of x2 – 3x + p = 0 and c, dare roots of x2 – 12x + q = 0, where a, b, cd, form a G.P. Prove that (q + p): (q – p) = 17:15.

Solution:

Given, a and b are the roots of x– 3p = 0

So, we have a + b = 3 and ab = p  … (i)

Also, c and d are the roots of x2 – 12x + q = 0

So, d = 12 and cd = q  … (ii)

And given abcd are in G.P.

Let’s take xb = xrc = xr2d = xr3

From (i) and (ii), we get

x + xr = 3

x (1 + r) = 3

And,

xr2 + xr3 =12

xr(1 + r) = 12

On dividing, we get

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 29

When r = 2, x = 3/(1 + 2) = 3/3 = 1

When r = -2, x = 3/(1 – 2) = 3/-1 = -3

Case I:

When r = 2 and x =1,

ab = x2r = 2

cd = x2r5 = 32

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 30

Case II:

When = –2, = –3,

ab = x2r = –18

cd = x2r5 = – 288

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 31

Therefore, in both the cases, we get (q + p): (q – p) = 17:15\

19. The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 32.

Solution:

Let the two numbers be a and b.

A.M = (a + b)/ 2 and G.M. = √ab

From the question, we have

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 33

20. If a, b, c are in A.P,; b, c, d are in G.P and 1/c, 1/d, 1/e are in A.P. prove that a, c, e are in G.P.

Solution:

Given abc are in A.P.

Hence, b – a = c – b  … (1)

And, given that bcd are in G.P.

So, c2 = bd  … (2)

Also, 1/c, 1/d, 1/e are in A.P.

So,

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 34

Now, required to prove that ace are in G.P. i.e., c2 = ae

From (1), we have

2b = a + c

b = (a + c)/ 2

And from (2), we have

d = c2/ b

On substituting these values in (3), we get

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 35

Therefore, ac, and e are in G.P.

21. Find the sum of the following series up to n terms:

(i) 5 + 55 + 555 + … (ii) .6 + .66 + . 666 + …

Solution:

(i) Given, 5 + 55 + 555 + …

Let Sn = 5 + 55 + 555 + ….. up to n terms

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 36

(ii) Given, .6 + .66 + . 666 + …

Let Sn = 06. + 0.66 + 0.666 + … up to n terms

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 37

22. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Solution:

Given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms

∴ nth term = an = 2n × (2n + 2) = 4n2 + 4n

The 20th term,

a20 = 4 (20)2 + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

Therefore, the 20th term of the series is 1680.

23. Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

Solution:

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ an–1 an

S = 3 + 7 + 13 + 21 + …. + an – 2 a– 1 + an

On subtracting both the equations, we get

S – S = [3 + (7 + 13 + 21 + 31 + …+ an–1 an)] – [(3 + 7 + 13 + 21 + 31 + …+ an–1) an]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – an

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – an

an = 3 + [4 + 6 + 8 + … (n –1) terms]

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 38

24. If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22 = S3 (1 + 8S1).

Solution:

From the question, we have

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 39

Therefore, from (1) and (2), we have 9S22 = S3 (1 + 8S1).

25. Find the sum of the following series up to n terms:NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 40

Solution:

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 41

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 42

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 43

26. Show that 

Solution:

nth term of the numerator = n(n + 1)2 = n3 + 2n2 + n

nth term of the denominator = n2(n + 1) = n3 + n2

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 44

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 45

27. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him?

Solution:

Given, the farmer pays Rs 6000 in cash.

So, the unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

From the question, the interest paid annually will be

12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500

Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)

= 12% of (500 + 1000 + 1500 + … + 6000)

It’s seen that, the series 500, 1000, 1500 … 6000 is an A.P. with the first term and common difference both equal to 500.

Let’s take the number of terms of the A.P. to be n.

So, 6000 = 500 + (n – 1) 500

1 + (n – 1) = 12

n = 12

Now,

The sum of the A.P = 12/2 [2(500) + (12 – 1)(500)] = 6 [1000 + 5500] = 6(6500) = 39000

Hence, the total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)

= 12% of 39000 = Rs 4680

Therefore, the tractor will cost the farmer = (Rs 12000 + Rs 4680) = Rs 16680

28. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Solution:

Given, Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

So, the unpaid amount = Rs 22000 – Rs 4000 = Rs 18000

Form the question, it’s understood that the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000

Hence, the total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000

= 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of (1000 + 2000 + 3000 + … + 18000)

It’s seen that, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000.

Let’s take the number of terms to be n.

So, 18000 = 1000 + (n – 1) (1000)

n = 18

Now, the sum of the A.P is given by:

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 51

Thus,

Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of Rs 171000 = Rs 17100

Therefore, the cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

Solution:

It’s seen that,

The numbers of letters mailed forms a G.P.: 4, 42, … 48

Here, first term = 4 and common ratio = 4

And the number of terms = 8

The sum of n terms of a G.P. is given by:

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 46

Also, given that the cost to mail one letter is 50 paisa.

Hence, Cost of mailing 87380 letters = Rs 87380 x (50/100) = Rs 43690 = Rs 43690

Therefore, the amount spent when 8th set of letter is mailed will be Rs 43690.

30. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

Solution:

Given, the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

Hence, the interest in first year = (5/100) x Rs 10000 = Rs 500 

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 47So, Amount in 15th year = Rs 

= Rs 10000 + 14 × Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 48And, the amount after 20 years = 

= Rs 10000 + 20 × Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Therefore, the amount in the 15th year is Rs 17000 and the total amount after 20 years will be Rs 20000.

31. A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

Solution:

Given, the cost of machine = Rs 15625

Also, given that the machine depreciates by 20% every year.

Hence, its value after every year is 80% of the original cost i.e., 4/5th of the original cost.

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 49Therefore, the value at the end of 5 years =  

= 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years will be Rs 5120.

32. 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.

Solution:

Let’s assume x to be the number of days in which 150 workers finish the work.

Then from the question, we have

150x = 150 + 146 + 142 + …. (x + 8) terms

The series 150 + 146 + 142 + …. (x + 8) terms is an A.P.

With first term (a) = 150, common difference (d) = –4 and number of terms (n) = (x + 8)

Now, finding the sum of terms:

NCERT Solutions Class 11 Mathematics Chapter 9 misc ex. - 50

As x cannot be negative. [Number of days is always a positive quantity]

x = 17

Hence, the number of days in which the work should have been completed is 17.

But, due to the dropping out of workers the number of days in which the work is completed

= (17 + 8) = 25


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

Enlisted beneath are the important concepts of Maths included in the Chapter 9 Sequences and Series:

9.1 Introduction

9.2 Sequences

9.3 Series

9.4 Arithmetic Progression (A.P.)

9.4.1 Arithmetic mean

9.5 Geometric Progression (G. P.)

9.5.1 General term of a G.P

9.5.2. Sum to n terms of a G.P

9.5.3 Geometric Mean (G.M.)

9.6 Relationship Between A.M. and G.M.

9.7 Sum to n Terms of Special Series

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

The chapter Sequences and Series belongs to the unit Algebra, that adds up to 30 marks of the total 80 marks. There are 4 exercises along with a miscellaneous exercise in this chapter to help students understand the concepts related to Sequences and Series clearly. Some of the topics discussed in Chapter 9 of NCERT Solutions for Class 11 Maths are as follows:

  1. By a sequence, we mean an arrangement of numbers in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3, ….k}. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence.
  2. Let a1, a2, a3, … be the sequence, then the sum expressed as a1 + a2 + a3 + … is called series. A series is called finite series if it has a finite number of terms.
  3. An arithmetic progression (A.P.) is a sequence in which terms increase or decrease regularly by the same constant. This constant is called the common difference of the A.P. Usually, we denote the first term of A.P. by a, the common difference by d and the last term by l. The general term or the nth term of the A.P. is given by an = a + (n – 1) d.
  4. A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the nth term of G.P. is given by an = arn – 1

A student who has mastered the chapter Sequences and Series of Class 11 would also have a strong hold on the concepts related to the chapter, namely, Sequence and Series, Arithmetic Progression (A. P.), Arithmetic Mean (A.M.), Geometric Progression (G.P.), general term of a G.P., sum of n terms of a G.P., infinite G.P. and its sum, geometric mean (G.M.) and the relation between A.M. and G.M.

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