Question

A. Find the sum of the following arithmetic series.

(1) 2 + 5 + 8 + ……. upto 12 terms

(2) 1 + 4 + 7 + ……..upto 22 terms

(3) 3 + 7 + 11 + …….upto 35 terms

(4) Find the sum of arithmetic series which contains 25 terms and whose middle term is 20.

B. In an A.P. if

(1) T_n = 4n + 3 find S₁₅

(2) T_n = 5 − 2n find S₂₀

C. (1) Find the sum of all the first ‘n’ odd Natural numbers.

(2) Find the sum of all Natural numbers between 101 and 201 which are divisible by 4.

Answer 1

(A) The sum of the first n terms of an A.P. is,

(1)

2, 5, 8 ... forms an A.P. with and

The sum of the given series up to 12 terms can be calculated as:

(2)

1, 4, 7 … forms an A.P. with and

The sum of the given series up to 22 terms can be calculated as:

(3)

3, 7, 11 … forms an A.P. with and

The sum of the given series up to 35 terms can be calculated as:

(4) It is given that n = 25 and the middle term is 20 i.e., thirteenth term, = 20.

Hence, the sum of the arithmetic series is 500.

(B)

(1) T_n = 4n + 3

(2) T_n = 5 − 2n

(C)

(1) The first n odd natural numbers are 1, 3, 5, 7, 9 … n

It can be observed that these numbers form an A.P. whose first term is 1 and common difference is 2 i.e., and .

Hence, the sum of all the first n odd natural numbers is .

(2) The list of all natural numbers between 101 and 201, which are divisible by 4 is:

104, 108 … 200

It can be observed that these numbers form an A.P. whose first term is 104, last term is 200 and common difference is 4 i.e., and.