Question

Find the three consecutive terms in an

a. p. whose sum is 18 and the sum of their squares 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

Answer 2

I hope my answer is help u ...