Question

The sum of four integers in A.P. is $24$, and their product is $945$; find them.

Answer 1

Let the integers that are in A.P are

$a-3d,a-d,a+d,a+3d$

Given

$a-3d+a-d+a+d+a+3d=24\Rightarrow 4a=24\Rightarrow a=6(a-3d)(a-d)(a+d)(a+3d)=945(a^2-{9d}^2)(a^2-d^2)=945(36-{9d}^2)(36-d^2)=945(4-d^2)(36-d^2)=105144-4d^2-36d^2+d^4=105d^4-40d^2+39=0d^4-d^2-39d^2+39=0d^2(d^2-1)-39(d^2-1)=0(d^2-39)(d^2-1)=0d^2=39,1$

But $d$ must be an integer

$\Rightarrow d^2=1\Rightarrow d=\pm 1$

So two sequences of number can be formed with $a=6$ and $d=\pm 1$

$9,7,5,3$ and $3,5,7,9$