If sum of the first 21 terms of the series
Question:

If sum of the first 21 terms of the series $\log _{9} 1 / 2 x+\log _{9}^{1 / 3} x+\log _{9} 1 / 4 x+\ldots .$, where $x>0$ is 504 , then $x$ is equal to

  1. 243

  2. 9

  3. 7

  4. 81


Correct Option: , 4

Solution:

$\mathrm{s}=2 \log _{9} \mathrm{x}+3 \log _{9} \mathrm{x}+\ldots \ldots+22 \log _{9} \mathrm{x}$

$\mathrm{s}=\log _{9} \mathrm{x}(2+3+\ldots . .+22)$

$\mathrm{s}=\log _{9} \mathrm{x}\left\{\frac{21}{2}(2+22)\right\}$

Given $252 \log _{9} x=504$

$\Rightarrow \log _{9} \mathrm{x}=2 \Rightarrow \mathrm{x}=81$

Administrator

Leave a comment

Please enter comment.
Please enter your name.