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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 to the product of two middle terms is 7: 15. Find the numbers.
Solution
Let the 4 numbers which are in AP be
a−3d, a−d, a+d, a+3d ...(i)
Given, the sum of these terms = 32
⇒(a−3d)+(a−d)+(a+d)+(a+3d)=32
⇒ 4a = 32
⇒ a = 8
Given, the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15
`((a - 3d)(a+ 3d))/((a-d)(a+d)) = 7/15`
`(a^2 - 9d^2)/(a^2 - d^2) = 7/15`
`15a^2 - 135d^2 = 7a^2 - 7d^2`
`8a^2 = 128d^2`
`128d^2 = 512` (∵ a = 8)
`d^2 = 4`
`d = +- 2`
Putting the value `d = +- 2` and a = 8 in (i) we get the requied number as 2,6,10,14
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