Question

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is \(\frac{27}{19}\). Then the common ratio of this series is :

• A. \(\frac{4}{9}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{2}{9}\)

Hint:

Write both sums in terms of a and r, eliminate a, and solve the resulting quadratic in r with the condition \(0 \lt r \lt 1\).

Answer 1

The Correct Option is C

Let the first term be \(a\) and ratio \(r\) (\(0\lt r\lt 1\) since terms are positive and the series converges).
\(\frac{a}{1-r}=3 \Rightarrow a = 3(1-r)\). Cubes form a GP with first term \(a^3\), ratio \(r^3\): \(\frac{a^3}{1-r^3}=\frac{27}{19}\).
\(\frac{27(1-r)^3}{(1-r)(1+r+r^2)}=\frac{27}{19} \Rightarrow 19(1-r)^2 = 1 + r + r^2\).
\(19 - 38r + 19r^2 = 1 + r + r^2 \Rightarrow 18r^2 - 39r + 18 = 0 \Rightarrow 6r^2 - 13r + 6 = 0\).
\(r = \frac{13 \pm 5}{12} = \frac{3}{2}\) or \(\frac{2}{3}\). Only \(r=\frac23\) satisfies \(|r|\lt1\).
Answer: (3) \(\frac{2}{3}\)