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JEE Mains · Maths · STD 11 - 8. sequence and series

\(60\) तथा \(n\) पदों की दो \(G.P.\) क्रमशः \(2,2^2, 2^3, \ldots\) तथा \(4,4^2, 4^3, \ldots\) हैं। यदि सभी \(60+ n\) पदों का गुणोत्तर माध्य \((2)\) \({ }^{\frac{225}{8}}\) है, तो \(\sum \limits_{ k =1}^{ n } k ( n - k )\) बराबर है :

  1. A \(560\)
  2. B \(1540\)
  3. C \(1330\)
  4. D \(2600\)
Verified Solution

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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