(13) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three terms.

Solution:
Let a – d , a, a + d are the first three terms.
Sum of three terms = 6
a – d + a + a + d = 6
                 3a = 6
                   a = 6/3
                   a = 2
Product of three terms = -120
(a – d) a (a + d) = -120
a (a²-d²) = - 120
2 (2² –d²) = -120
2(4 - d²) = -120
 (4 - d²) = -120/2
  (4 - d²) = -60
   - d² = -60 – 4
   - d² = -64
           d = √64
        d = ± 8
a = 2 d = 8      a = 2 d = -8
-6 , 2 , 10           10,2,-6

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