JEE Mains · Maths · STD 11 - 8. sequence and series
The sum of first four terms of a geometric progression \((G.P.)\) is \(\frac{65}{12}\) and the sum of their respective reciprocals is \(\frac{65}{18} .\) If the product of first three terms of the \(G.P.\) is \(1,\) and the third term is \(\alpha\), then \(2 \alpha\) is ....... .
- A \(5\)
- B \(6\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
Let number are \(a , ar , ar ^{2}, ar ^{3}\) \(a \frac{\left(r^{4}-1\right)}{r-1}=\frac{65}{12}......(1)\) \(\frac{1}{a} \frac{\left(\frac{1}{r^{4}}-1\right)}{\frac{1}{r}-1}=\frac{65}{18}\) \(\frac{1}{a r^{3}}\left(\frac{1-r^{3}}{1-r}\right)=\frac{65}{18}......(2)\)…
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