The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

The sum of infinite terms of a $G.P.$ is $x$ and on squaring each term of it,the sum becomes $y$. Then the common ratio of this series is:

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \leq 0$,then:

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 $a_{1}, a_{2}, a_{3}, \ldots$ be a $G$.$P$. such that $a_{1} < 0$; $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$. If $\sum_{i=1}^{9} a_{i} = 4 \lambda$,then $\lambda$ is equal to:

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of the squares of the first three terms is $33033$,then the sum of these three terms is equal to