Sum

Find five numbers in G.P. such that their product is 243 and sum of second and fourth number is 10.

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#### Solution

Let the five numbers in G.P. be

`"a"/"r"^2, "a"/"r"`, a, ar, ar^{2}.

According to the first condition,

`"a"/"r"^2 xx "a"/"r" xx "a" xx "ar" xx "ar"^2` = 243

∴ a^{5} = 243

∴ a = 3

According to the second condition,

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

∴ `1/"r" + "r" = 10/"a"`

∴ `(1 + "r"^2)/"r" = 10/3`

∴ 3r^{2} – 10r + 3 = 0

∴ 3r^{2} – 9r – r + 3 = 0

∴ (3r – 1) (r – 3) = 0

∴ r = `1/3, 3`

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

`"a"/"r"^2 = 27, "a"/"r" = 9, "a" = 3, "ar" = 1, "ar"^2 = 1/3`

When a = 3, r = 3

`"a"/"r"^2 = 1/3, "a"/"r" = 1, "a" = 3, "ar" = 9, "ar"^2 = 1/3`

∴ the five numbers in G.P. are

27, 9, 3, 1, `1/3 or 1/3`, 1, 3, 9, 27.

Concept: Sequence and Series - Geometric Progression (G.P.)

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