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
$m/n$
$\sqrt {mn} $
$mn$
$0$
Insert three numbers between $1$ and $256$ so that the resulting sequence is a $G.P.$
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 $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a $G.P.$ Prove that $P ^{2} R ^{n}= S ^{n}$
Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0 .$ If $\alpha$ $\beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is
Let ${A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ..... + {\left( { - 1} \right)^{n - 1}}{\left( {\frac{3}{4}} \right)^n}$ and $B_n \,= 1 - A_n$ . Then, the least odd natural number $p$ , so that ${B_n} > {A_n}$, for all $n \geq p$ is