Question

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

Answer 1

Let the four consecutive numbers in AP be $(a-3d),(a-d),(a+d)$ and $(a+3d)$

So$,$ according to the question$.$

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

$4a=32$

$a=32/4$

$a=8......\left(1\right)$

Now$,$ $(a-3d)(a+3d)/(a-d)(a+d)=7/15$

$15(a²-9d²)=7(a²-d²)$

$15a²-135d²=7a²-7d²$

$15a²-7a²=135d²-7d²$

$8a²=128d²$

Putting the value of $a=8$ in above we get$.$

$8\left(8\right)²=128d²$

$128d²=512$

$d²=512/128$

$d²=4$

$d=2$

So$,$ the four consecutive numbers are

$8-(3\times 2)$

$8-6=2$

$8-2=6$

$8+2=10$

$8+(3\times 2)$

$8+6=14$

Four consecutive numbers are $2,6,10and14$