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

यदि श्रेणी \(\log _{9^{1 / 2}}  x +\log _{9^{1 / 3}}  x +\log _{9^{1 / 4}} x +\ldots ., x >0\) के प्रथम \(21\) पदों का योग \(504\) है, तो \(x\) बराबर है

  1. A \(81\)
  2. B \(243\)
  3. C \(7\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(81\)

Step-by-step Solution

Detailed explanation

\(s=2 \log _{9} x+3 \log _{9} x+\ldots+22 \log _{9} x\) \(s=\log _{9} \times(2+3+\ldots+22)\) \(s=\log _{9} x\left\{\frac{21}{2}(2+22)\right\}\) Given \(252\,\log _{9} x=504\) \(\Rightarrow \log _{9} x=2 \Rightarrow x=81\)
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