The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds these term by 6, find three terms.
Let the three terms be a - d, a, a + d
a – d + a + a – d = 21 (given)
3a = 21
a = 7
According to question,
(a – d) (a + d) = a + 6
a2 – d2 = a + 6
72 – d2 = 7 + 6
49 – d2 = 13
d2 = 36
d = 6
Therefore, a – d = 7 – 6 = 1
a + d = 7 + 6 = 13
Thus, the three terms are 1, 7 and 13.