**NCERT
Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1**

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series
Ex 9.1 are the part of NCERT Solutions for Class 11 Maths. Here you can find
the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1.

**Write the first five terms of each of the sequences in Exercises 1 to 6
whose n ^{th} terms are:**

**Ex 9.1 Class 11 Maths Question 1.**

a_{n}= n(n + 2)

**Solution:**

Given, a_{n} = n(n + 2)

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

a_{1} = 1(1 + 2) = 1 × 3 = 3

a_{2} = 2(2 + 2) = 2 × 4 = 8

a_{3} = 3(3 + 2) = 3 × 5 = 15

a_{4} = 4(4 + 2) = 4 × 6 = 24

a_{5} = 5(5 + 2) = 5 × 7 = 35

∴ The first five terms of the give sequence are 3,
8, 15, 24 and 35, respectively.

**Ex 9.1
Class 11 Maths Question 2.**

a_{n}= n/(n+1)

**Solution:**

**Ex 9.1 Class 11 Maths Question 3.**

a_{n}= 2

^{n}

**Solution:**

**Ex 9.1
Class 11 Maths Question 4.**

a_{n}= (2n−3)/6

**Solution:**

**Ex 9.1 Class 11 Maths Question 5.**

a_{n}= (-1)

^{n-1}5

^{n+1}

**Solution:**

Given, a_{n} = (-1)^{n-1} 5^{n+1}

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

a_{1} = (-1)^{1-1} 5^{1+1} = (-1)^{°} 5^{2} =
25

a_{2} = (-1)^{2-1} 5^{2+1} = (-1)^{1} 5^{3} =
-125

a_{3} = (-1)^{3-1} 5^{3+1} = (-1)^{2} 5^{4} =
625

a_{4} = (-1)^{4-1} 5^{4+1} = (-1)^{3} 5^{5} =
-3125

a_{5} = (-1)^{5-1} 5^{5+1} = (-1)^{4} 5^{6} =
15625

∴ The first five terms of the given sequence are 25,
-125, 625, -3125 and 15625, respectively.

**Ex 9.1
Class 11 Maths Question 6.**

a_{n}= n(n

^{2}+ 5)/4

**Solution:**

Given, a_{n} = n(n^{2} + 5)/4

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

**Find the
indicated terms in each of the sequences in Exercises 7 to 10 whose n ^{th}
terms are:**

**Ex 9.1 Class 11 Maths Question 7.**

a_{n}= 4n – 3; a

_{17}, a

_{24}

**Solution:**

Given, a_{n} = 4n – 3

Substituting n = 17, we get a_{17} = 4(17) – 3 = 68 – 3 = 65

Substituting n = 24, we get a_{17 }= 4(24) – 3 = 96 – 3 = 93

**Ex 9.1
Class 11 Maths Question 8.**

a_{n}= n

^{2}/2

^{n}; a

_{7}

**Solution:**

Given, a_{n} = n^{2}/2^{n }

Substituting n = 7, we get a_{7 }= (7)^{2}/2^{7}
= 49/128^{}

**Ex 9.1 Class 11 Maths Question 9.**

a_{n}= (-1)

^{n – 1}n

^{3}; a

_{9}

**Solution:**

Given, a_{n} = (-1)^{n – 1} n^{3}^{}

Substituting n = 9, we get a_{9 }= (-1)^{9 – 1} 9^{3}
= (-1)^{8} × 729 = 729

**Ex 9.1
Class 11 Maths Question 10.**

a_{n}= n(n−2)/(n+3); a

_{20}

**Solution:**

Given, a_{n} = n(n−2)/(n+3)

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

**Ex 9.1 Class 11 Maths Question 11.**

a_{1}= 3, a

_{n}= 3a

_{n-1 }+ 2 for all n > 1

**Solution:**

Given, a_{1} = 3, a_{n} = 3a_{n-1 }+ 2

⇒ a_{1} = 3, a_{2} = 3a_{1} +
2 = 3 × 3 + 2 = 9 + 2 = 11,

a_{3} = 3a_{2} + 2 = 3 × 11 + 2 = 33 + 2 = 35,

a_{4} = 3a_{3} + 2 = 3 × 35 + 2 = 105 + 2 = 107,

a_{5} = 3a_{4} + 2 = 3 × 107 + 2 = 321 + 2 = 323

Hence, the first five terms of the given sequence are 3, 11, 35, 107 and 323,
respectively.

The corresponding series is 3 + 11 + 35 + 107 + 323 + ………..

**Ex 9.1
Class 11 Maths Question 12.**

a_{1}= -1, a

_{n}= a

_{n−1}/n, n ≥ 2

**Solution:**

Given,

The corresponding series is:

**Ex 9.1 Class 11 Maths Question 13.**

a_{1}= a

_{2}= 2, a

_{n}= a

_{n-1}– 1, n > 2

**Solution:**

Given, a_{1} = a_{2} = 2, a_{n} = a_{n-1} –
1, n > 2

a_{1} = 2, a_{2} = 2, a_{3 }= a_{2} –
1 = 2 – 1 = 1,

a_{4} = a_{3} – 1 = 1 – 1 = 0 and a_{5} =
a_{4} – 1 = 0 – 1 = -1

Hence, the first five terms of the given sequence are 2, 2, 1, 0 and -1,
respectively.

The corresponding series is 2 + 2 + 1 + 0 + (-1) + ……

**Ex 9.1
Class 11 Maths Question 14.**

The Fibonacci sequence is defined by 1 = a_{1}= a

_{2}and a

_{n}= a

_{n-1}+ a

_{n-2}, n > 2.

Find a

_{n+1}/a

_{n}, for n = 1, 2, 3, 4, 5

**Solution:**

We have,

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