The number of terms of the G.P. 3, 3 2 , 3 4 , needed to obtain a sum of 3069 512 is
Mathematics · JEE Main · NTA Exams — Progression and Series
The number of terms of the G.P. \(3, \frac{3}{2}, \frac{3}{4}, \ldots\) needed to obtain a sum of \(\frac{3069}{512}\) is
- \(n = 8\)
- \(n = 9\)
- \(n = 10\)
- \(n = 11\)
Answer
(C) n = 10
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