Question

The sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15. find the numbers.

Answer 1

the numbers are 2,6,10,14 or 14,10,6,2.

Answer 2

↑ Here is your answer ↓
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The sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15. find the numbers.

Four consecutive numbers in an AP are (a - 3d), (a - d), (a + d) & (a + 3d)

$a - 3d + a - d + a + d + a + 3d = 32$

'd' will cancel out,

$4a = 32$

$a = 8$

$(a-3d)(a+3d)/(a+d)(a-d) = 7/15$

$a^2 - 9d^2 / a^2 - d^2 = 7/15$

$15a^2 - 135d^2 = 7a^2 - 7d^2$

$8a^2 - 128d^2 = 0$

Divide throughout by 8,

$a^2 - 16d^2 = 0$

$a^2 = 16d^2$

Now substitute 'a',

$8^2 = 16d^2$

$64 = 16d^2$

$d^2 = 64/16 = 4$

$d = \sqrt{4} = +2 (or) -2$

$a - 3d = 8 - 3(2) = 8 - 6 = 2$

$a - d = 8 - 2 = 6$

$a + d = 8 + 2 = 10$

$a + 3d = 8 + 3(2) = 8 + 6 = 14$

∴ $AP =\ \textgreater \ 2, 6, 10, 14...$

(or) $14, 10, 6, 2...$ (if d = -2)

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Glad to help you.
@EmadAhamed