Alg8.7

The sum of first three terms of an A.P is $36\,$ while their product is $1620\,$ .Find the A.P.

Let the three terms be $a-d,a,a+d\,$

Given that $a-d+a+a+d=36,3a=36,a=12\,$

Also $(a-d)a(a+d)=1620\,$

$a(a^2-d^2)=1620\,$

$12(144-d^2)=1620,144-d^2=135\,$

$d^2=9,d=\pm3\,$

We therefore have two arithmetic progressions with two values of d

First one is for $d=3\,$

$9,12,15\,$

Next one is $d=-3\,$

$15,12,9\,$