JEE Mains · Maths · STD 11 - 8. sequence and series
If sum of the first \(21\) terms of the series \(\log _{9^{1 / 2}} x +\log _{9^{1 / 3}} x +\log _{9^{1 / 4}} x +\ldots ., x >0\) , where \(x>0\) is \(504,\) then \(\mathrm{x}\) is equal to:
- A \(81\)
- B \(243\)
- C \(7\)
- D \(9\)
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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