JEE Mains · Maths · STD 11 - 8. sequence and series
Consider two G.Ps. \(2,2^{2}, 2^{3}, \ldots\) and \(4,4^{2}, 4^{3}, \ldots\) of \(60\) and \(n\) terms respectively. If the geometric mean of all the \(60+n\) terms is \((2)^{\frac{225}{8}}\), then \(\sum_{ k =1}^{ n } k (n- k )\) is equal to.
- A \(560\)
- B \(1540\)
- C \(1330\)
- D \(2600\)
Answer & Solution
Correct Answer
(C) \(1330\)
Step-by-step Solution
Detailed explanation
\(\left(\left(2^{1} 2^{2} \cdots 2^{60}\right)\left(4^{1} \cdot 4^{2} \ldots \ldots 4^{ n }\right)\right)^{\frac{1}{60+ n }}=2^{\frac{225}{8}}\) \(\left(2^{30 \times 61} 4^{\frac{ n ( n +1)}{2}}\right) \frac{1}{60+ n }=2^{\frac{225}{8}}\)…
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