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
$=6[0.1+0.11+0.111+\ldots . \text { to } n \text { terms }]$
$=\frac{6}{9}[0.9+0.99+0.999+\ldots . . \text { to } n \text { terms }]$
$=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)+\ldots . \text { to } n \text { terms }\right]$
$=\frac{2}{3}\left[(1+1+\ldots n \text { terms })-\frac{1}{10}\left(1+\frac{1}{10}+\frac{1}{10^{2}}+\ldots n \text { terms }\right)\right]$
$=\frac{2}{3}\left[n-\frac{1}{10}\left(\frac{1-\left(\frac{1}{10}\right)^{n}}{1-\frac{1}{10}}\right)\right]$
$=\frac{2}{3} n-\frac{2}{30} \times \frac{10}{9}\left(1-10^{-n}\right)$
$=\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$
If $a$,$b$,$c \in {R^ + }$ are such that $2a$,$b$ and $4c$ are in $A$.$P$ and $c$,$a$ and $b$ are in $G$.$P$., then
If $x,\,2x + 2,\,3x + 3,$are in $G.P.$, then the fourth term is
A $G.P.$ consists of an even number of terms. If the sum of all the terms is $5$ times the sum of terms occupying odd places, then find its common ratio.
Find the sum to indicated number of terms in each of the geometric progressions in $\left.x^{3}, x^{5}, x^{7}, \ldots n \text { terms (if } x \neq\pm 1\right)$