Question

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

Answer 1

Let the numbers be: $a-d,a,a+d$

Given:

$a-d+a+a+d=21$

$3a=21$

$\therefore a=7$

$(a-d)(a+d)=a+6⟹d^2=a^2-a-6$

Put $a=7$, we get

$d^2=36⟹d=6$

Therefore the numbers are: $a-d,a,a+d=7-6,7,7+6=\{1,7,13\}$