If the continued product of three numbers in G.P. is 216 and the sum o

1.	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» 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

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»

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