Find the sum of the following series up to n terms:
$6+.66+.666+\ldots$
$6+.66+.666+\ldots$
Let $S_{n}=06+0.66+0.666+\ldots .$ to $n$ terms
$=6[0.1+0.11+0.111+\ldots . \text { to } n \text { terms }]$
$=\frac{6}{9}[0.9+0.99+0.999+\ldots . . \text { to } n \text { terms }]$
$=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)+\ldots . \text { to } n \text { terms }\right]$
$=\frac{2}{3}\left[(1+1+\ldots n \text { terms })-\frac{1}{10}\left(1+\frac{1}{10}+\frac{1}{10^{2}}+\ldots n \text { terms }\right)\right]$
$=\frac{2}{3}\left[n-\frac{1}{10}\left(\frac{1-\left(\frac{1}{10}\right)^{n}}{1-\frac{1}{10}}\right)\right]$
$=\frac{2}{3} n-\frac{2}{30} \times \frac{10}{9}\left(1-10^{-n}\right)$
$=\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$
The sum of a $G.P.$ with common ratio $3$ is $364$, and last term is $243$, then the number of terms is
If $1 + \cos \alpha + {\cos ^2}\alpha + .......\,\infty = 2 - \sqrt {2,} $ then $\alpha ,$ $(0 < \alpha < \pi )$ is
If $1\, + \,\sin x\, + \,{\sin ^2}x\, + \,...\infty \, = \,4\, + \,2\sqrt 3 ,\,0\, < \,x\, < \,\pi $ then
Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b .$
The sum of infinity of a geometric progression is $\frac{4}{3}$ and the first term is $\frac{3}{4}$. The common ratio is