**Exercise 8.1**

Sequence and Series

**Question and Answers**

**Class 11 – Maths**

Sequence and Series

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Sequences and Series |

Chapter No. | Chapter 8 |

Exercise | Exercise 8.1 |

Category | Class 11 Maths NCERT Solutions |

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

**(1) a _{n} = n (n + 2)**

**Answer** n^{th} term of a sequence a_{n} = n (n + 2)** **

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

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.

**(2) a _{n} = n/n+1**

**Answer** Given the n^{th} term, a_{n} = n/n+1

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

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

**(3) a_{n} = 2^{n}**

**Answer** Given the n^{th} term, *a _{n}* = 2

^{n}On substituting* n* = 1, 2, 3, 4, 5, we get

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 32

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

**(4) a _{n} = (2n – 3)/6**

**Answer** Given the n^{th} term, *a _{n}* = (2n – 3)/6

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

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

**(5) a _{n} = (-1)^{n-1} 5^{n+1}**

**Answer** Given the n^{th} term, a_{n} = (-1)^{n-1} 5^{n+1}

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

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

**(6) **

**Answer** On substituting *n* = 1, 2, 3, 4, 5, we get the first 5 terms.

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 n ^{th} terms are:**

**(7) a _{n} = 4n – 3; a_{17}, a_{24}**

**Answer **The* n*^{th} term of the sequence is a_{n} = 4n – 3

On substituting *n* = 17, we get

a_{17} = 4(17) – 3 = 68 – 3 = 65

Next, on substituting *n* = 24, we get

a_{24} = 4(24) – 3 = 96 – 3 = 93

**(8) a _{n} = n^{2}/2^{n} ; a^{7}**

**Answer ** The* n*^{th} term of the sequence is a_{n} = n^{2}/2^{n}

Now, on substituting *n* = 7, we get

a_{7} = 7^{2}/2^{7} = 49/ 128

**(9) a _{n} = (-1)^{n-1} n^{3}; a_{9}**

**Answer ** The* n*^{th} term of the sequence is a_{n} = (-1)^{n-1} n^{3}

On substituting *n* = 9, we get

a_{9} = (-1)^{9-1} (9)^{3} = 1 x 729 = 729

**(10) a _{n} = n (n^{2} + 5 ) /4 ; a_{20}**

**Answer **On substituting *n* = 20, we get

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

**(11) a _{1} = 3, a_{n} = 3a_{n-1} + 2 for all n > 1**

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

Then,

a_{2} = 3a_{1} + 2 = 3(3) + 2 = 11

a_{3} = 3a_{2} + 2 = 3(11) + 2 = 35

a_{4} = 3a_{3} + 2 = 3(35) + 2 = 107

a_{5} = 3a_{4} + 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) a _{1} = -1, a_{n} = a_{n-1}/n, n ≥ 2**

**Answer ** a_{n} = a_{n-1}/n and a_{1} = -1

Then,

a_{2} = a_{1}/2 = -1/2

a_{3} = a_{2}/3 = -1/6

a_{4} = a_{3}/4 = -1/24

a_{5} = a_{4}/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) a _{1} = a_{2 }= 2, a_{n} = a_{n-1} – 1, n > 2**

**Answer ** a_{1} = a_{2}, a_{n} = a_{n-1} – 1

Then,

a_{3} = a_{2} – 1 = 2 – 1 = 1

a_{4} = a_{3} – 1 = 1 – 1 = 0

a_{5} = a_{4} – 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 = 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**Answer** 1 = a_{1} = a_{2}

a_{n} = a_{n – 1 }+ a_{n – 2}, n > 2

So,

a_{3} = a_{2} + a_{1} = 1 + 1 = 2

a_{4} = a_{3} + a_{2} = 2 + 1 = 3

a_{5} = a_{4} + a_{3} = 3 + 2 = 5

a_{6} = a_{5} + a_{4} = 5 + 3 = 8

Thus,

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