Question

The sum and the product of the three concecutive numbers in a geometric progression are 31 and 125 respectively. Find the three numbers of the G.P.

Answer 1

Suppose, the three consecutive numbers In a GP. are a/r, a, ar.

The sum of the three consecutive terms = 28

∴ a/r + a + ar = 28

∴ \(\frac{a+ar+ar^2}{r}\) = 28

∴ a + ar + ar² = 28r ………..(1)

The product of the three consecutive temrs = 512

∴ a/r × a × ar = 512

∴ a³ = 512 = (8)³

∴ a = 8

Putting a = 8 In the result 1,

8 + 8r + 8r² = 28r

∴ 8r²+ 8r - 28r+80

∴ 8r² – 20r + 8 = 0

∴ 2r² – 5r + 2 = 0

∴ 2r² – 4r – r + 2 = 0

∴ 2r(r – 2) – 1(r – 2)=O

∴ (r – 2)(2r – 1) = 0

∴ r – 2 = 0 OR 2r – 1 = 0

∴ r = 2 OR r = 1/2

Putting a = 8 and r = 2 in a/r, a, ar,

Three consecutive numbers, we get

8/2, 8, 8 × 2 ⇒ 4, 8, 16

Putting a = 8 and r = 1/2 In a/r, a, ar,

Three consecutive numbers we get are :

8/1/2, 8, 8 × 1/2 ⇒ 16, 8, 4.

Hence, three consecutive numbers are

4, 8, 16 OR 16, 8, 4.