Question

Find three positive integers in A.P. such that their sum is 24 and their product is 480.

Answer 1

Let the three terms be a - d, a, a + d
Sum = 24 Product = 480
$\therefore a-\text{/}d+a+a+\text{/}d=24$
$3a=24$
$\therefore a=8$
$(a-d)\left(a\right)(a+d)=480$
$(8-d)\left(8\right)(8+d)=480[\because a=8]$
$(8-d)(8+d)=60$
$64-d^2=60$
$d^2=\pm \sqrt{4}=\pm 2$
If $a=8,d=2$ the terms are
$a-d=8-2=6,a=8,a+d=8+2=10$
$\therefore$ the term are $6,8,10.$

We will get the same values if we'll consider $d=-2$.