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 is 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,

        a-3d+a-d+a+d+a+3d=32

        4a=32

        a=8 1

        (a-3d)(a=3d) / (a-d)(a+d)=7/15

        15(a2-9d2) = 7(a2-d2)

         15a2-135d2 = 7a2-7d2

          8a2-128d2 = 0

          d2 = 4, ±2

Therefore d=4  or d=±2 1mark

So, when a=8 and d=2, the numbers are 2,6,10,14.

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

Answer 2

let the
four numbers are.
a-3d,a-d,a+d,a+3d
add them all
and here we get
4a=32
a=8
and as given..
a+3d×a-3d/a+d×a-d=7/15
from here we get
a=8
so the A. P series is
2,6,10,14
hope it helps u.