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 some terms of $G.P.$ is $315$ whose first term and the common ratio are $5$ and $2,$ respectively. Find the last term and the number of terms.
Insert two numbers between $3$ and $81$ so that the resulting sequence is $G.P.$
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
Find the sum of first $n$ terms and the sum of first $5$ terms of the geometric
series $1+\frac{2}{3}+\frac{4}{9}+\ldots$
Find the sum of the following series up to n terms:
$5+55+555+\ldots$