Question

The sum and the product of the three consecutive terms of G.P. are 6 and -64 respectively. Find the three terms of the G.P.

Answer 1

Suppose, three consecutive terms of G.P. are

ar, a, ar. r

The sum of the three consecutive terms = 6

∴ a/r + a + ar = 6 .

\(\frac{a+ar+ar^2}{r}\) = 6

a + ar + ar² = 6r ………….(1)

The product of the three consecutive terms = -64

∴ a/r × a × ar = – 64

a³ = – 64 = (- 4)³

∴ a = – 4

Putting a = -4 in the result (1),

– 4 – 4r – 4r² = 6r

4r² + 4r + 6r + 4 = 0

4r² + 10r + 4 = 0

∴ 2r² + 5r + 2 = 0

∴ 2r² + 4r + r + 2 = 0

∴ 2r(r + 2) + 1 (r + 2) = 0

(r + 2) (2r + 1) = 0

r + 2 = 0 OR 2r + 1 = 0

∴ r = – 2 OR r = \(-\frac{1}{2}\)

Putting a = – 4 and r = – 2 in a/r; a; a/r,

Three consecutive terms:

\(\frac{−4}{−2}\), -4, (-4) (- 2} ⇒ 2, -4, 8.

Putting a = – 4 and r = 1/2 in y; a; y

Three consecutive terms :

\(\frac{−4}{−2}\), -4, (-4)(\(-\frac{1}{2}\)) ⇒ 8, -4, 2.

Hence, three consecutive terms are 2, -4, 8 OR 8, -4, 2.