Question

\(a_{1}, a_{2}, \ldots , a_{10}\) are in G.P., and if \(a_{1} + a_{2} = 6, a_{9} + a_{10} = \frac{3}{128}\) then the common ratio of the G.P. is equal to

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

Hint:

In GP, \(a_{n+k} = a_n r^k\). Use ratio of sums to eliminate \(a_1\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
\(a_1 + a_2 = a_1(1+r) = 6\). \(a_9 + a_{10} = a_1 r^8 (1+r) = \frac{3}{128}\).

Step 2: Detailed Explanation:
Divide second by first: \(r^8 = \frac{3/128}{6} = \frac{3}{128 \cdot 6} = \frac{1}{256} = (\frac{1}{2})^8\). So \(r = 1/2\).

Step 3: Final Answer:
Option (B).