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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 + ... $ upto $20$ terms is $840$, then $a$ is equal to
Sequences and Series
Solution:
Given expression can be written as
$2log_{2}a + 4log_{2}a + 6log_{2}a + ... + 40log_{2}a$
$ = log_{2}a \left\{2 + 4 + ... + 40\right\}$
$ = 10\left(2 + 40\right) log_{2}a$
$ = 420 \,log_{2}a$
Now, $420\, log_{2}a = 840$
$\Rightarrow log_{2}a = 2$
$ \Rightarrow a = 4$.