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Question
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is `57/2`. Find the three terms
Solution
Let the three consecutive terms in a G.P. are `"a"/"r"`, a, ar.
Their Product = `"a"/"r" xx "a" xx "ar"` = 27
a3 = 27 = 33
a = 3
Sum of the product of terms taken two at a time is `57/2`
`"a"/"r" xx "a" + "a" xx "ar" + "ar" xx "a"/"r" = 57/2`
`"a"^2/"r" + "a"^2"r" + "a"^2 = 57/2`
`3^2(1/"r" + "r" + 1) = 57/2`
`(1 + "r"^2 + "r")/"r" = 57/2 xx 1/9 = 57/18`
18 + 18r2 + 18r = 57r
18r2 + 18r – 57r + 18 = 0
18r2 – 39r + 18 = 0 ÷ 3
⇒ 6r2 – 13r + 6 = 0
`("r" - 2/3)("r" - 3/2)` = 0
r = `2/3, 3/2`
If a = 3, r = `2/3`
∴ The three numbers is `3/(2/3), 3, 3 xx 2/3`
or `3 xx 2/3, 3, 3 xx 2/3`
`9/2, 3, 2`
If a = 3, r = `3/2`, the three numbers is
`"a"/"r", "a", "ar" = 3/(3/2), 3, 3 xx 3/2`
= `6/3, 3, 9/2`
= `2, 3, 9/2`
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