Sum of four consecutive number in an AP is 32 and the ratio of products of first and last term to the product of middle term is 7:15. Find the no.

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Solution

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
d^2 = 4,
d = ±2

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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Sum of four consecutive number in an AP is 32 and the ratio of products of first and last term to the product of middle term is 7:15. Find the no.

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 d^2 = 4, d = ±2 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

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
d^2 = 4,
d = ±2

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