Question:medium
\(a_{1}, a_{2}, \ldots , a_{10}\) 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
\(a_{1}, a_{2}, \ldots , a_{10}\) 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
Show Hint
In GP, \(a_{n+k} = a_n r^k\). Use ratio of sums to eliminate \(a_1\).
Updated On: Apr 25, 2026
- \(\frac{1}{3}\)
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- 2
- 3
Show Solution
The Correct Option is B
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