Question

The sum of four consecutive numbers in A.P. is $32$ and the ratio of the product of the first and last terms to the product of two middle terms is $7:15$ . Find the number .

Answer 1

Let the four terms of the AP be $a-3d,a-d,a+d$ and $a+3d$.
Given:
$(a-3d)+(a-d)+(a+d)+(a+3d)=32$
$\Rightarrow 4a=32$
$\Rightarrow a=8$
Also, according to the question,

$\Rightarrow \frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{7}{15}$

$\Rightarrow \frac{8^2-9d^2}{8^2-d^2}=\frac{7}{15}$

$\Rightarrow \frac{64-9d^2}{64-d^2}=\frac{7}{15}$

$\Rightarrow 15(64-9d^2)=7(64-d^2)$

$\Rightarrow 960-135d^2=448-7d^2$

$\Rightarrow 128d^2=512$

$\Rightarrow d^2=4$

$\Rightarrow d=\pm 2$

When $a=8$ and $d=2$, then the terms are $2,6,10,14$.

When $a=8$ and $d=-2$, then the terms are $14,10,6,2$.