Question

Five positive integer are in $A.P$ the sum of middle $3$ term is $24$ and product of first and fifth term $48.$ Find the term of $A.P$

Answer 1

Series be $a-2d,a-d,a,a+d,a+2d$

The sum of first $3$ terms

$a-d+a+a+d=24$

$3a=24$

$a=24÷3$

$a=8$

$(a-2d)(a+2d)=48$

$a^2-4d^2=48$

substitute $a=8$

$64-4d^2=48$

$4d^2=64-48$

$d^2=16$

$d=2$

Therefore, series are $4,6,8,10,12$