Question

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

Answer 1

Let,the numbers in A.P be $a-3d,a-d,a+d,a+3d$

Given: Sum of four consecutive numbers $=32$

$\therefore a-3d+a-d+a+d+a+3d=32$

$⟹4a=32$

$⟹a=\frac{32}{4}=8$

Now,

A.T.Q,

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

$⟹\frac{a^2-9d^2}{a^2-d^2}=\frac{7}{15}$

$⟹\frac{64-9d^2}{64-d^2}=\frac{7}{15}[\because a=8]$

$⟹(64-9d^2)\times 15=7(64-d^2)$

$⟹64\times 15-135d^2=7\times 64-7d^2$

$⟹64\times 15-7\times 64=135d^2-7d^2$

$⟹64(15-7)=128d^2$

$⟹64\times 8=128d^2$

$⟹\frac{64\times 8}{128}=d^2$

$⟹d^2=5$

$⟹d=\pm 2$

$\therefore$ Either , the numbers are $2,6,10,14$

or

$14,10,6,2$