Question

The sum of four consecutive terms which are in an arithmetic progression 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 4 consecutive numbers in AP be

$(a-3d),(a-d),(a+d),(a+3d)$

According to the question,

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

$4a=32$

$a=8$

Now,

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

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

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

$8a^2=128d^2$

Putting the value of a, we get,

$d^2=4$

$d=\pm 2$

So, the four consecutive numbers are $2,6,10,14$ or $14,10,6,2$.