Let the three consecutive terms in an A.P. be m – d,m,m + d
By the given data,
Sum of three terms = 18
m – d + m + m + d = 18
3m = 18
m = 18/6 = 6
Again by the given data,
Sum of their squares = 140
(m – d)2 + m2 + (m + d)2 = 140
m2 + d2 – 2md + m2 + m2 + d2 + 2md = 140
3m2 + 2d2 = 140
3(62) + 2d2 = 140
3(36) + 2d2 = 140
2d2 = 140 – 108
2d2 = 32
d2 = 32/2 = 16
∴ d = ± 4
when, m = 6, d = + 4
m – d = 6 – 4 = 2
m = 6
m + d = 6 + 4 = 10
when, m = 6, d = -4
m – d = 6-(-4) = 6 + 4= 10
m = 6
m + d = 6 +(-4) = 6 – 4 = 2
∴ The three numbers are 2, 6 and 10 or 10, 6,2
