Answer:

Given.

The sum of first three terms is 21

Assume, the first three terms as a – d, a, a + d [where a is the first term and d is the common difference]

Sum of first three terms is

a – d + a + a + d = 21

3a = 21

a = 7

It is also given that product of first and third term exceeds the second by 6

So, (a – d)(a + d) – a = 6

a² – d² – a = 6

Substituting the value of a = 7, we get

7² – d² – 7 = 6

d² = 36

d = 6 or d = – 6

Hence, the terms of AP are a – d, a, a + d which is 1, 7, 13.