Question

The sum of four consecutive number in AP is 32 and the ratio of the product of the first and the last term to the product of two middle tem is 7:5.find the numbers

Answer 1

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

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Answer 2

RATIO IS 7:15

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.