How many terms of the $G.P.$ $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the sum $\frac{3069}{512} ?$
Let $n$ be the number of terms needed. Given that $a=3, r=\frac{1}{2}$ and $S_{n}=\frac{3069}{512}$
Since $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
Therefore $\frac{3069}{512}=\frac{3\left(1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}}=6\left(1-\frac{1}{2^{n}}\right)$
or $\frac{3069}{3072}=1-\frac{1}{2^{n}}$
or $\frac{1}{2^{n}}=1-\frac{3069}{3072}=\frac{3}{3072}=\frac{1}{1024}$
or $2^{n}=1024=2^{10},$ which gives $n=10$
If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then
In a $G.P.,$ the $3^{rd}$ term is $24$ and the $6^{\text {th }}$ term is $192 .$ Find the $10^{\text {th }}$ term.
The sum of infinite terms of a $G.P.$ is $x$ and on squaring the each term of it, the sum will be $y$, then the common ratio of this series is
Let ${a_n}$ be the ${n^{th}}$ term of the G.P. of positive numbers. Let $\sum\limits_{n = 1}^{100} {{a_{2n}}} = \alpha $ and $\sum\limits_{n = 1}^{100} {{a_{2n - 1}}} = \beta $, such that $\alpha \ne \beta $,then the common ratio is
The $G.M.$ of the numbers $3,\,{3^2},\,{3^3},....,\,{3^n}$ is