Question

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

Answer 1

Let the 3rd term of A.P. be
a - d, a, a + d
Then,
a - d + a + a + d = 21
$\Rightarrow 3a=21$
$\therefore a=7$
and
$\Rightarrow (a-d)(a+d)=a+6\Rightarrow a^2-d^2=a+6$
$\Rightarrow 7^2-a^2=7+6$ $[\because a=7]$
$\Rightarrow d^2=36\Rightarrow d=\pm 6$
When d = 6, a = 7, we get : 1, 7, 13
When d = -6, a = 7, we get : 13, 7, 1