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  4. If log 21 / 2 a+ log 21 / 4 a+ log 21 / 6 a+ log 21 / 8 a+ ldots. upto 20 terms is 840 , then a equal to

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Q. If $\log _{2^{1 / 2}} a+\log _{2^{1 / 4}} a+\log _{2^{1 / 6}} a+\log _{2^{1 / 8}} a+\ldots$. upto 20 terms is 840 , then a equal to

Sequences and Series

Solution:

$=2 \log _2 a +4 \log _2 a +6 \log _2 a +\ldots \ldots \ldots \ldots+40 \log _2 a $
$=\log _2 a \{2+4+\ldots \ldots+40\}$
$=\frac{20}{2}(2+40) \log _2 \text { a } $
$=420 \log _2 a =840 $
$\Rightarrow \log _2 a =2$
$\Rightarrow a=4$