Sum

Find five numbers in G. P. such that their product is 1024 and fifth term is square of the third term.

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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 given conditions,

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

∴ a^{5} = 4^{5}

∴ a = 4 ...(i)

Also, ar^{2} = a^{2}

∴ r^{2} = a

∴ r^{2} = 4 ...[From (i)]

∴ r = ±2

When a = 4, r = 2

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

When a = 4, r = – 2

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

∴ the five numbers in G.P. are

1, 2, 4, 8, 16 or – 2, 4, – 8, 16.

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

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