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प्रश्न
Find three consecutive numbers in G.P. whose sum is 28 and product is 512.
बेरीज
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उत्तर
`a/r`, a, ar
`a/r` + a + ar = 28
`a(1/r + 1 + r)` = 28
`a/r xx a xx ar` = 512
⇒ a3 = 512
a = 8
`8(1/r + 1 + r)` = 28
2(1 + r + r2) = 7r
2r2 + 2r − 7r + 2 = 0
2r2 − 5r + 2 = 0
2r2 − 4r − r + 2 = 0
2r(r − 2) − 1(r − 2) = 0
(2r − 1)(r − 2) = 0
r = `1/2`, 2
1. r = `1/2`, a = 8
= `8/(1/2), 8, 8(1/2)`
= 16, 8, 4
2. r = 2, a = 8
`8/2, 8, 8(2)`
= 4, 8, 16
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