Question

Find three numbers in G.P., such that their sum is 35 and their product is 1000.

Answer 1

Let the three numbers in G.P. be `"a"/"r"`, a, ar.

According to the first condition,

`"a"/"r" + "a" + "ar"` = 35

∴ `"a"(1/"r" + 1 + "r")` = 35          ...(i)
According to the second condition,

`("a"/"r")("a")("ar")` = 1000

∴ a³ = 1000
∴ a = 10
Substituting the value of a in (i), we get
`10(1/"r" + 1 + "r")` = 35

∴ `1/"r" + "r" + 1 = 35/10`

∴ `1/"r" + "r" = 35/10 - 1`

∴ `1/"r" + "r" = 25/10`

∴ `1/"r" + "r" = 5/2`

∴ 2r² – 5r + 2 = 0
∴ (2r – 1) (r – 2) = 0

∴ r = `1/2 or "r" = 2`

When r = `1/2, "a" = 10`

`"a"/"r" = 10/((1/2)) = 20, "a" = 10 and  "ar" = 10(1/2)` = 5

When r = 2, a = 10

`"a"/"r" = 10/2` = 5, a = 10 and ar = 10(2) = 20

∴ the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.