Question -
Answer -
Given:
The sum of first threeterms is 21
Let us assume thefirst three terms as a – d, a, a + d [where a is the first term and d is thecommon difference]
So, sum of first threeterms is
a – d + a + a + d = 21
3a = 21
a = 7
It is also given thatproduct of first and third term exceeds the second by 6
So, (a – d)(a + d) – a= 6
a2 – d2 –a = 6
Substituting the valueof a = 7, we get
72 – d2 –7 = 6
d2 =36
d = 6 or d = – 6
Hence, the terms of APare a – d, a, a + d which is 1, 7, 13.