## NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.1

Chapter 9 Sequence and Series Exercise 9.1 NCERT Solutions for Class 11 Maths is very important topic which you need to improve your marks in the examinations. Here you will find Class 11 Maths NCERT Solutions that will help you a lot in understanding the basics things of the chapter and implementing them. It will save your precious time if you want to complete your homework on time.

1. Write the first five terms of each of the sequence defined by the following:a

_{n}Â = n (n + 2)

**Answer**

a

_{n}Â = n (n + 2)

For n = 1, a

_{1}Â = 1(1 + 2) = 3

For n = 2, a

_{2}Â = 2(2 + 2) = 8

For n = 3, a

_{3}Â = 3(3 + 2) = 15

For n = 4, a

_{4}Â = 4(4 + 2) = 24

For n = 5, a

_{5}Â = 5(5 + 2) = 35

Thus first five terms are 3, 8, 15, 24, 35.

2. Write the first 5 terms of the sequence defined by the following:

a

_{n}Â = n/(n + 1)

**Answer**

a

_{n}Â = n/(n + 1)

for n = 1, a

_{1}Â = 1/(1 + 1) = 1/2

for n = 2, a

_{2}Â = 2/(2 +1 ) = 2/3

for n = 3, a

_{3}Â = 3/(3 + 1) = 3/4

for n = 4, a

_{4}Â = 4/(4 + 1) = 4/5

for n = 5, a

_{5}Â = 5/(5 + 1) = 5/6

Hence, first 5 terms of the given sequence are

1/2, 2/3, 3/4, 4/5 and 5/6.

3. Write the first five terms of the sequence whose n

^{th}Â term is: a

_{n}Â = 2

^{n}

**Answer**

a

_{n}Â = 2

^{n}, Putting n = 1, 2, 3, 4, 5

a

_{1}Â = 2

^{1}Â = 2, a

_{2}Â = 2

^{2}Â = 4, a

_{3}Â = 2

^{3}Â = 8

a

_{Â 4}Â = 2

^{4}Â = 16, a

_{5}Â = 2

^{5}Â = 32

Required first five term of sequences are 2, 4, 8, 16, 32.

4. Write the first five terms of the sequence whose n

^{Â th}Â term is: a

_{n}Â = (2n â€“ 3)/6.

**Answer**

Here a

_{n}Â = (2n â€“ 3)/6

Putting n = 1, 2, 3, 4, 5, we get

a

_{1}Â = (2Â Ã—Â 1 â€“ 3)/6 = (2 â€“ 3)/6 = -(1/6);

a

_{2}Â = (2 Ã—Â 2 â€“ 3)/6 = (4 â€“ 3)/6 = 1/6;

a

_{3}Â = (2 Ã—Â 3 â€“ 3)/6 = (6 â€“ 3)/6 = 3/6 = 1/2;

a

_{4}Â = (2 Ã—Â 4 â€“ 3)/6 = (8 â€“ 3)/6 = 5/6

and a

_{5}Â = (2 Ã—Â 5 â€“ 3)/6 = (10 â€“ 3)/6 = 7/6

âˆ´Â the first five terms are â€“(1/6), 1/6, 1/2, 5/6 and 7/6.

5. Write the first five terms of the sequence whose n

^{Â th}Â term is a

_{n}Â = (â€“1)

^{n â€“ 1}Â 5

^{Â n + 1Â }

**Answer**

Putting n = 1, 2, 3, 4, 5

a

_{1}Â = (â€“1)

^{0}. 5Â

^{1 + 1}Â = 5

^{2}Â = 25

a

_{2}Â = (â€“1)

^{1}. 5Â

^{2 + 1}Â = â€“ 5

^{3}Â = â€“125

a

_{3}Â = (â€“1)

^{2}. 5Â

^{3 + 1}Â = 5

^{4}Â = 625

a

_{4}Â = (â€“1)

^{3}. 5

^{Â 4 + 1Â }= â€“ 5

^{5}Â = â€“ 3125

a

_{5}Â = (â€“1)

^{4}. 5

^{Â 5 + 1Â }= 5

^{6}Â = 15625

6. Write the first 5 terms of the sequence defined by the following:

a

_{n}Â = (n(n

^{2}Â + 5))/4

**Answer**

for n = 1, a

_{1}Â = (1(1

^{2}Â + 5))/4 = 6/4 = 3/2

for n =2 , a

_{2}Â = (2(2

^{2}Â + 5))/4 = (2Â Ã—Â 9)/4 = 9/2

for n = 3, a

_{3}Â = (3(3

^{2}Â + 5))/4 = (3 Ã—Â 14)/4 = 21/2

for n = 4, a

_{4}Â = (4(4

^{2}Â + 5))/4 = (4 Ã—Â 21)/4 = 21

for n = 5, a

_{5}Â = (5(5

^{2}Â + 5))/4 = (5 Ã—Â 30)/4 = 75/2

hence the first 5 terms are 3/2, 9/2, 21/2, 21 and 75/2

7. Find the indicated terms in the following sequence whose n

^{Â th}Â term is:

a

_{n}Â = 4n â€“ 3, a

_{17}, a

_{24}

**Answer**

a

_{n}Â = 4n â€“ 3

For n = 17, a

_{17}Â = 4 Ã— 17 â€“ 3 = 68 â€“ 3 = 65

For n = 24, a

_{24}Â = 4 Ã— 24 â€“ 3 = 96 â€“ 3 = 93

Hence a

_{17}Â = 65 and a

_{24}Â = 93

8. Find the indicated terms in the following sequence whose n

^{Â th}Â term is:

a

_{n}Â = n

^{2}/2

^{n}Â ; a

_{7}

**Answer**

a

_{n}Â = n

^{2}/2

^{n}

putting n = 7

a

_{7}Â = 7

^{2}/2

^{7}Â = 49/128

9. Find the indicated term in the following sequence whose n

^{Â th}Â term is:

a

_{n}Â = (â€“1)

^{n â€“ 1}Â n

^{Â 3}, a

_{9}

**Answer**

a

_{n}Â = (â€“1)

^{n â€“ 1}Â n

^{3}, Putting n = 9

a

_{9}Â = (â€“1)

^{9 â€“ 1}Â 9

^{3}Â = 729

10. Find the indicated terms in the sequence whose n

^{Â th}Â term is a

_{n}Â = (n(n -2))/(n + 3) ; a

_{20}

**Answer**

a

_{n}Â = (n(n -2))/(n + 3),

Putting n = 20,

a

_{20}Â = (20(20 â€“ 2))/(20 + 3) = (20Â Ã—Â 18)/23 = 360/23

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

_{1}Â = 3, a

_{n}Â = 3, a

_{n â€“ 1Â }+ 2 for all n > 1.

**Answer**

a

_{1}Â = 3, a

_{n}Â = 3 a

_{n â€“1Â }+ 2. Putting n = 2, 3, 4, 5

a

_{2}Â = 3.a

_{1}Â + 2 = 3. 3 + 2 = 9 + 2 = 11

a

_{3}Â = 3.a

_{2}Â + 2 = 3. 11 + 2 = 33 + 2 = 35

a

_{4}Â = 3.a

_{3}Â + 2 = 3. 35 + 2 = 105 + 2 = 107

a

_{5}Â = 3.a

_{4}Â + 2 = 3. 107 + 2 = 321 + 2 = 323

The first five terms of the sequences are

3, 11, 35, 107, 323

âˆ´Â The corresponding series is

3 + 11 + 35 + 107 + 323 +....

12. a

_{1}Â = -1, a

_{n}Â = (a

_{n -1})/n , nÂ â‰¥Â 2

**Answer**

Here a

_{1}Â = -1, a

_{n}Â = (a

_{n â€“ 1})/n , nÂ â‰¥ 2

Putting n = 2, 3, 4, 5 and 6

a

_{2}Â = (a

_{2}Â â€“ 1)/2 = a

_{1}/2 = -1/2

a

_{3}Â = (a

_{3}Â â€“ 1)/3 = a

_{2}/2 = -(1/2)/3 = -1/6;

a

_{4}Â = (a

_{4}Â â€“ 1)/4 = a

_{3}/4 = -(1/6)/4 = -1/24;

a

_{5}Â = a

_{5}-1/5 = a

_{4}/5 = -(1/24)/5 = -1/120

The first five terms of the sequences are

-1, -(1/2), -(1/6), -(1/24), -1/120

The corresponding series is

-1, -(1/2), -(1/6), -(1/24), -1/120 â€¦..

13. a

_{1}Â = a

_{2}Â = 2, a

_{n}Â = a

_{nâ€“ 1Â }â€“ 1, n > 2.

**Answer**

Given: a

_{1}Â = a

_{2}Â = 2

and a

_{n}Â = a

_{nâ€“ 1Â }â€“ 1, n > 2.

For n = 3, a

_{3}Â = a

_{2 â€“ 1Â }= 2 â€“ 1 = 1

For n = 4, a

_{4}Â = a

_{3 â€“ 1Â }= 1 â€“ 1 = 0

For n = 5, a

_{5}Â = a

_{4 â€“ 1Â }= 0 â€“ 1 = â€“1

Hence first five terms are 2, 2, 1, 0, â€“1 and the corresponding series = 2 + 2 + 1 + 0 + (â€“1) +....

14. The Fibonacci sequence is defined by

1 = a

_{1}Â = a

_{2}Â and an = a

_{n â€“ 1Â }+ a

_{n â€“ 2}, n > 2

Find (a

_{n+ 1})/a

_{n}Â , for n = 1, 2, 3, 4, 5

**Answer**

For n = 1, (a

_{n+1})/a

_{n}Â = a

_{2}/a

_{1}Â = 1/1 = 1 a

_{1}Â = a

_{2}Â = 1

and a

_{n}Â = a

_{n â€“ 1Â }+ a

_{n â€“ 2}, n > 2 â€¦(A)

n = 3 in eqn (A) a

_{3}Â = a

_{2}Â + a

_{1}Â = 1 + 1 = 2

n = 4 in eqn (A) a

_{4}Â = a

_{3}Â + a

_{2}Â = 2 + 1 = 3

n = 5 in eqn (A) a

_{5}Â = a

_{4}Â + a

_{3}Â = 3 + 2 = 5

n = 6 in eqn (A) a

_{6}Â = a

_{5}Â + a

_{4}Â = 5 + 3 = 8

for n = 2, (a

_{n+1})/a

_{n}Â = a

_{3}/a

_{2}Â = 2/1 = 2;

for n = 3, (a

_{n+1})/a

_{n}Â = a

_{4}/a

_{3}Â = 3/2;

for n = 4, (a

_{n+1})/a

_{n}Â = a

_{5}/a

_{4}Â = 5/3;

for n = 5, (a

_{n+1})/a

_{n}Â = a

_{6}/a

_{5}Â = 8/5

The values of (a

_{n+1})/a

_{n}Â = for n = 1, 2, 3, 4, 5 are 1, 2, 3/2 , 5/3 , 8/5