Question

the sum of four consecutive numbers in an AP is 32 , and the ratio of the product of first and last term 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,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

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Answer 2

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