The sum of four consecutive numbers in an ap is 32 and the ratio of the product of the first and last term to the product of 2 middle terms is 7:15. find the number
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The sum of four consecutive no. in AP is 32 and the ratio of the product of
the first and last term to the product of two middle term is 7:15. Find
no.
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Let the no/s be of the form (a-3d) (a-d) (a+d) (a+2d)
Then according to question,
(a-3d)+(a-d)+(a+d)+(a+3d) = 32
4a = 32
a = 8
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(a-3d)(a+3d) / (a-d)(a+d) = 7/15
15(a²-9d²) = 7(a²-d²)
15a² - 135d² = 7a² - 7d²
8a² - 128d² = 0
⇒Divide by 8 on whole
⇒ a² - 16d²
⇒Substituting a =8, Then
⇒ 8² - 16d²
⇒64 - 16d²
⇒(8+4d)(8-4d)
Then d = +2, -2
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If d = +2 , Then
No/s - (2, 6, 10,14)
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If d = -2 , Then
No/s - (14,10, 6, 2)
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Hence the no/s would be (2, 6, 10, 14) or (14, 10, 6, 2)
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