Sum

Find four numbers in G. P. such that sum of the middle two numbers is `10/3` and their product is 1.

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#### Solution

Let the four numbers in G.P. be `"a"/"r"^3, "a"/"r"`, ar, ar^{3}.

According to the second condition,

`"a"/"r"^3 ("a"/"r") ("ar")("ar"^3)` = 1

∴ a^{4} = 1

∴ a = 1

According to the first condition,

`"a"/"r" + "ar" 10/3`

∴ `1/"r" + (1)"r" = 10/3`

∴ `(1 + "r"^2)/"r" = 10/3`

∴ 3 + 3r^{2 }= 10r

∴ 3r^{2} – 10r + 3 = 0

∴ (r – 3)(3r – 1) = 0

∴ r = 3 or r = `1/3`

When r = 3, a = 1

`"a"/"r"^3 = 1/(3)^3 = 1/27, "a"/"r" = 1/3, "ar"` = 1(3) = 3 and ar^{3} = 1(3)^{3} = 27

When r = `1/3`, a = 1

`"a"/"r"^3 = 1/((1/3)^3) = 27, "a"/"r" = 1/((1/3)) = 3`,

`"ar" = 1(1/3) = 1/3 and "ar"^3 = 1(1/3)^3 = 1/27`

∴ the four numbers in G.P.are

`1/27, 1/3, 3, 27 or 27, 3, 1/3, 1/27`.

Is there an error in this question or solution?

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