If ${\log _x}a,\;{a^{x/2}}$ and ${\log _b}x$ are in $G.P.$, then $x = $
$ - \log ({\log _b}a)$
$ - {\log _a}({\log _a}b)$
${\log _a}({\log _e}a) - {\log _a}({\log _e}b)$
${\log _a}({\log _e}b) - {\log _a}({\log _e}a)$
If the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a $G.P.$ are $a, b$ and $c,$ respectively. Prove that
$a^{q-r} b^{r-p} c^{p-q}=1$
Let $a_{n}$ be the $n^{\text {th }}$ term of a G.P. of positive terms.
If $\sum\limits_{n=1}^{100} a_{2 n+1}=200$ and $\sum\limits_{n=1}^{100} a_{2 n}=100,$ then $\sum\limits_{n=1}^{200} a_{n}$ is equal to
If ${(p + q)^{th}}$ term of a $G.P.$ be $m$ and ${(p - q)^{th}}$ term be $n$, then the ${p^{th}}$ term will be
The sum of the first three terms of a $G.P.$ is $S$ and their product is $27 .$ Then all such $S$ lie in
Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to