Find the sum to $n$ terms of the sequence, $8,88,888,8888 \ldots$
The given sequence is $8,88,888,8888 \ldots$
This sequence is not a $G.P.$ However, it can be changed to $G.P.$ by writing the terms as
$S_{n}=8+88+888+8888+\ldots \ldots$ to $n$ terms
$=\frac{8}{9}[9+99+999+9999+\ldots \ldots . . $ to $ n $ terms $]$
$=\frac{8}{9}[(10+10^{2}+\ldots \ldots . n \text { terms })$
$-(1+1+1+\ldots . . n \text { terms })]$
$=\frac{8}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]$
$=\frac{8}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]$
$=\frac{80}{81}\left(10^{n}-1\right)-\frac{8}{9} n$
The sum of first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$ Find the common ratio and the terms.
If $x,\;y,\;z$ are in $G.P.$ and ${a^x} = {b^y} = {c^z}$, then
Find the sum of the following series up to n terms:
$6+.66+.666+\ldots$
If $a,\;b,\;c$ are in $A.P.$, then ${10^{ax + 10}},\;{10^{bx + 10}},\;{10^{cx + 10}}$ will be in
The difference between the fourth term and the first term of a Geometrical Progresssion is $52.$ If the sum of its first three terms is $26,$ then the sum of the first six terms of the progression is