Solution:

Let a – d , a, a + d are the first three terms.

Sum of three terms = 18

a – d + a + a + d = 18

3a = 18

a = 18/3

a = 6

Sum of their squares = 140

(a – d)² + a² + (a + d)² = 140

a² + d² – 2ad + a² + a² + d² + 2ad = 140

3 a² + 2 d² = 140

3(6)² + 2 d² = 140

3(36) + 2d² = 140

108 + 2d² = 140

2d² = 140 – 108

2d²= 32

d²= 32/2

d²= 16

d = √16

d = ± 4

d = 4 d = -4

a = 6 d = 4 a = 6 d = -4

2,6,10 10,6,2

Therefore the three terms are 2,6,10 or 10,6,2