Question

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

Answer 1

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"/cancel("r"^2) xx "a"/cancel("r") xx "a" xx "a"cancel("r") xx "a"cancel("r"^2)` = 1024

∴ a⁵ = 4⁵

∴ a = 4           ...(i)

Also, ar² = a²

∴ `"r"^2 = a^2/a`

∴ 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 1, - 2, 4, - 8, 16.