Question

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

Answer 1

Suppose that,

The four consecutive number of an A.P be

$a,a+d,a+2d$ and $a+3d$

Given that,

$a+a+d+a+2d+a+3d=32$

$4a+6d=32$

$2a+3d=\frac{32}{2}$

$2a+3d=16$

$a=\frac{16-3d}{2}$

According to given question

$\frac{a(a+3d)}{(a+d)(a+2d)}=\frac{7}{15}$

$\Rightarrow 15a(a+3d)=7(a+d)(a+2d)$

$\Rightarrow 15a^2+45ad=7a^2+21ad+14d^2$

$\Rightarrow 8a^2+24ad-14d^2=0$

$\Rightarrow 4a^2+12ad-7d^2=0$

$\Rightarrow 4a^2+14ad-2ad-7d^2=0$

$\Rightarrow 2a(2a+7d)-d(2a+7d)=0$

$\Rightarrow (2a+7d)(2a-d)=0$

$\Rightarrow a=-\frac{7d}{2},a=\frac{d}{2}$

Substitute value of $a$ in equation (1) and we get,

When, $a=-\frac{7d}{2}$ and $a=\frac{d}{2}$

$\frac{-7d}{2}=\frac{16-3d}{2}\frac{d}{2}=\frac{16-3d}{2}$

$-7d=16-3dd=16-3d$

$-4d=164d=16$

$d=-4d=4$

Here

$d=4\left(positive\right)$

$then$

$a=\frac{d}{2}$

$a=\frac{4}{2}=2$

$a=2$

Therefore four consecutive number are

$a,a+d,a+2d,a+3d$

$2,6,10and14$

Hence, this is the answer.