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 ratio 15 find the numbers

Answer 1

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Answer 2

The answer is given below :

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

Given that,

(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

or,

14, 10, 6, 2.

Therefore, the four consecutive numbers are

2, 6, 10, 14.

Thank you for your question.