How many consecutive odd integers beginning with 5 will sum to 480?

Solution :

Consecutive integers starting from 5 are 5, 7, 9, ............

5 + 7 + 9 + 11 + ............. = 480

Sn = 480

(n/2)[2a + (n - 1)d] = 480

a = 5, d = 7 - 5 = 2

(n/2)[2(5) + (n - 1)(2)] = 480

(n/2)[10 + 2n - 2] = 480

(n/2)[2n + 8] = 480

2n2 + 8n = 960

Dividing by 2, we get

n2 + 4n - 480 = 0

(n + 24) (n - 20) = 0

n = -24 or n = 20

Hence by finding the sum of 20 terms we will get 480.