Question

The sum of 4 even consecutive positive integers of an AP is 68. The ratio of the product of first and last term to the product of two middle terms is 35:36. Find the numbers?​

Answer 1

let a - 3d, a - d, a + d, a + 3d be the 4 terms in A.P.

$a - 3d + a - d + a + d + a + 3d = 68 \\ 4a = 68 \\ a = 17 \\ {a}^{2} = 289 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i) \\ \\ now \\ \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{35}{36} \\ \frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{35}{36} \\ 36 {a}^{2} - 324 {d}^{2} = 35 {a}^{2} - 35 {d}^{2} \\ 36 {a}^{2} - 35 {a}^{2} = 324 {d}^{2} - 35 {d}^{2} \\ {a}^{2} = 289 {d}^{2} \\ 289 = 289 {d}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(from \: i) \\ {d}^{2} = 1 \\ d = 1$

Hence, the terms are

a - 3d = 17 - 3(1) = 17 - 3 = 14

a - d = 17 - 1 = 16

a + d = 17 + 1 = 18

a + 3d = 17 + 3(1) = 17 + 3 = 20

Hence, the terms of the given A.P. are 14, 16, 18 and 20.