(14) The sum of three consecutive terms in an A.P whose sum is 18 and the sum of their squares is 140.

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