Find the sum of the following A.P. 1, 3, 5, 7, ............, 199.

Given:

a = 1

d = 3 – 1 = 2

a _n = 199

We know that,

a _n = a + (n – 1)d

⇒ 199 = 1 + (n – 1)2

⇒ 199 – 1 = 2n – 2

⇒ 199 – 1 + 2 = 2n

⇒ 2n = 200

Now sum of n terms of AP,

⇒ S _n = 50 [2 + 99 × 2]

⇒ S _n = 50 [2 + 198]

⇒ S _n = 50 × 200

⇒ S _n = 10000