Question

If the continued product of three numbers in G.P. is 216 and the sum of their products in pair is 156, find numbers.

Answer 1

Let three numbers in G.P. be \(\frac{a}{r}\), a, ar … (i)

Given, \(\frac{a}{r}\). a. ar = 216

⇒ a³ = 216 = 6³

⇒ a = 6

Therefore sum of their product in pair are:

\(\frac{a}{r}\) × a + a × ar × \(\frac{a}{r}\) = 156

⇒ \(\frac{36}{r}\) + 36r + 36 = 156                 (∴ a = 6)

⇒ \(\frac{36}{r}\) + 36r = 120

⇒ \(\frac{3}{r}\) + 3r = 10

⇒ 3r² − 10r + 3 = 3

⇒ (3r − 1)(r − 3) = 0

⇒ r = \(\frac{1}{3}\), 3

Substituting value of a and r in eq (i), we get

18, 6, 2 when r = \(\frac{1}{3}\)

and 2, 6,18 when when r = 3