Find the sum of first $n$ terms and the sum of first $5$ terms of the geometric
series $1+\frac{2}{3}+\frac{4}{9}+\ldots$
Here $a=1$ and $r=\frac{2}{3} .$ Therefore
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{2}{3}}=3\left[1-\left(\frac{2}{3}\right)^{n}\right]$
In particular, $S_{5}=3\left[1-\left(\frac{2}{3}\right)^{5}\right]=3 \times \frac{211}{243}=\frac{211}{81}$
Find the sum of the sequence $7,77,777,7777, \ldots$ to $n$ terms.
The sum of $3$ numbers in geometric progression is $38$ and their product is $1728$. The middle number is
If ${x_r} = \cos (\pi /{3^r}) - i\sin (\pi /{3^r}),$ (where $i = \sqrt{-1}),$ then value of $x_1.x_2.x_3......\infty ,$ is :-
If $\frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}},\left( {x \ne 0} \right)$ then $a$, $b$, $c$, $d$ are in
If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then