Question

The sum of four integers which form an A.P. , is $24$ and their product is $945$. Find the terms of the A.P.

• A. $3,5,7,9$
• B. $1,15,9,7$
• C. $7,9,3,5$
• D. $15,9,1,7$

Answer 1

The correct option is A $3,5,7,9$

Let the four integers be $a-3d,a-d,a+d,a+3d$ which form an A.P.
Since, their sum is $24$
Therefore, $a-3d+a-d+a+d+a+3d=24$
$\Rightarrow a=6$
and their product is $945$ with $a=6$
Therefore, $(6-3d)(6-d)(6+d)(6+3d)=945$
$\Rightarrow 9d^4-360d^2+351=0$
Therefore, integral root is $d=1$
Therefore, terms of A.P are $6-3,6-1,6+1,6+3=3,5,7,9$
Ans: A