Q24 of 30 Page 1

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.

Let the numbers be : (a – 3d), (a – d), (a + d), (a + 3d)

(Note the common difference is “2d”)


According to Question,


• (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ a = 8


So, the numbers are: (8 – 3d), (8 – d), (8 + d), (8 + 3d)


Also,


•


⇒


(Using (a + b)(a – b) = a2 – b2)


⇒ 128d2 = 512


⇒ d2 = 4


⇒ d = ±2


So, putting d = 2, numbers are: 2, 6, 10, 14


(Note: putting d = – 2 gives the same numbers but in reverse order)


Numbers are 2, 6, 10, 14

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