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Arithmetic Progressions

Question
CBSEENMA10009710

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 numbers be (a -3d),(a-d), (a+d) and (a+3d)

∴ (a-3d) + (a-d) +(a+ d) + (a +3d) = 32

⇒ 4a = 32

a = 8

Also, (a-3d)(a+ 3d)(a-d)(a+d) =715 15a2 -135d2 = 7a2 -7d2 8a2 = 128d2d2 = 8a2128 = 8 x 8x8128d2 =4d =±2

If d =2 numbers are : 2, 6,10, 14
If d = -2 numers are 14,10,16,2