Question

Find three consecutive terms in ap whose sum is 18 and sum of their squars is 140

Answer 1

let the terms be a-d,a,a+d

a-d+a+a-d=18

3a=18

a=6

(a-d)^2+a^2+(a+d)^2=140

a^2-2ad+d^2+a^2+a^2+2ad+d^2=140

3a^2+2d^2=140

3(6)^2 + 2d^2=140

3(36)+2d^2=140

108+2d^2=140

2d^2=140-108

2d^2=32

d^2=16

d=4

so the terms are

(a-d),a,(a+d)

(6-4),6,(6+4)

2, 6, 10