Find five numbers in g.p. such that their product is 32 and the product of the last two is 108. *

Let the five numbers in G. P. be

`"a"/"r"^2, "a"/"r", "a", "ar", "ar"^2`

According to the given conditions,

`"a"/"r"^2 xx "a"/"r" xx "a" xx "ar" xx "ar"^2` = 1024

∴ a5 = 45∴ a = 4                   ...(i)

Also, ar2 = a2


∴ r2 = a
∴ r2 = 4            ...[From (i)]∴ r = ±2

When a = 4, r = 2

`"a"/"r"^2 = 1, "a"/"r" = 2`, a = 4, ar = 8, ar2 = 16

When a = 4, r = – 2

`"a"/"r"^2 = 1, "a"/"r" =  – 2`, a = 4, ar = – 8, ar2 = 16

∴  the five numbers in G.P. are
1, 2, 4, 8, 16 or – 2, 4, – 8, 16.

Let the five numbers in G.P. be

`"a"/"r"^2, "a"/"r"`, a, ar, ar2.

According to the first condition,

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

∴ a5 = 243∴ a = 3

According to the second condition,

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

∴ `1/"r" + "r" = 10/"a"`

∴ `(1 + "r"^2)/"r" = 10/3`

∴ 3r2 – 10r + 3 = 0
∴ 3r2 – 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.

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