Question

the sum of 4 consecutive number 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

Answer 1

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Answer 2

The answer is given below :

[For four number in AP, it is better to assume the numbers as (a - 3d), (a - d), (a + d) and (a + 3d).]

Let, the four consecutive numbers are
(a - 3d), (a - d), (a + d) and (a + 3d).

Given that, there sum :

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

=> 4a = 32

=> a = 8

So, the numbers are

(8 - 3d), (8 - d), (8 + d) and (8 + 3d).

Given that :

(8 - 3d)(8 + 3d) : (8 - d)(8 + d) = 7 : 15

=> (64 - 9d²) : (64 - d²) = 7 : 15

=> (64 - 9d²)/(64 - d²) = 7/15

=> 960 - 135d² = 448 - 7d²

=> 128d² = 512

=> d² = 4

So, d = ± 2.

So, the numbers are :

2, 6, 10, 14 (when d = 2)

or,

14, 10, 6, 2 (when d = -2)

Therefore, the four consecutive numbers are

2, 6, 10, 14.

You can also the sequence as

14, 10, 6, 2.

Thank you for your question.