Question

The sum of four integers in A.P is 24 and their product is 945. Find the numbers.

Answer 1

Let the $4$ integers be $(a-3d),(a-d),(a+d)$ and $(a+3d)$ respectively.

Now according to the question-

Sum of these integers is $24$.

$\therefore (a-3d)+(a-d)+(a+d)+(a+3d)=24$

$\Rightarrow a-3d+a-d+a+d+a+3d=24$

$\Rightarrow 4a=24\Rightarrow a=6$

Product of these numbers is $945$.

$\therefore (a-3d)(a-d)(a+d)(a+3d)=24$

$\Rightarrow (a^2-{\left(3d\right)}^2)(a^2-d^2)=945$

$\Rightarrow a^4-10a^2{d}^2+9d^4=945$

Substituting $a=6$, we have $\left[\text{From}\left(1\right)\right]$

$9d^4-360d^2+1296=945$

$\Rightarrow 9d^4-360d^2+351=0$

$\Rightarrow d^4-40d^2+39=0$

$\Rightarrow (d^2-39)(d^2-1)=0$

$\Rightarrow d^2=39,1$

Case I:-

$d^2=39$

$d=\sqrt{39}=6.24\left(\text{not possible}\right)$

Case II:-

$d^2=1$

$d=\sqrt{1}=1$

$\therefore a=6\text{and}d=1$

Hence the four integers are $3,5,7$ and $9$.