Question

The sum of $4$ 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 number.

Answer 1

Let the four consecutive numbers in AP be $a-3d,a-d,a+d,a+3d.$
So, Sum of $4$ consecutive numbers $=32$

$\Rightarrow (a-3d)+(a-d)+(a+d)+(a+3d)=32$

$\Rightarrow 4a=32$

$\Rightarrow a=8\dots \dots \left(i\right)$

Ang, it is also given that

$\frac{1^{st}term\times 4^{th}term}{2^{nd}term\times 3^{rd}term}=\frac{7}{15}$

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

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

$\Rightarrow 15a^2-135d^2=7a^2-7d^2$

$\Rightarrow 8a^2-128d^2=0$

$\Rightarrow (8{)}^2-16d^2=0$ [Using $eq.\left(i\right)$]

$\Rightarrow d^2=\frac{64}{16}=4$

$\Rightarrow d=\pm 2$

Therefore $d=\pm 2$
So, when $a=8$ and $d=2,$ the numbers are

$a-3d=8-3\times 2=2$

$a-d=8-2=6$

$a+d=8+2=10$

$a+3d=8+3\times 2=14.$

Similarly, When $a=8,d=-2$ the numbers are $14,10,6,2.$