Sum To n Terms Of a GP - Practice Questions & MCQ

The sum of the n-term of a GP

Let $S_n$ be the sum of $n$ terms of the G.P. with the first term ' $a$ ' and common ratio ' $r$ '. Then

$S_n=a+ar+a{r}^2+\dots +a{r}^{n-2}+a{r}^{n-1}$
Multiply both sides with $r$

$r{S}_n=ar+a{r}^2+a{r}^3+\dots +a{r}^{n-1}+a{r}^n$
Subtract (ii) from (i)

$\begin{array}{rl}& S_n-r{S}_n=a-a{r}^n\\ & \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}-{\mathrm{ar}}^{\mathrm{n}}}{1-\mathrm{r}}=\mathrm{a}\left(\frac{1-{\mathrm{r}}^{\mathrm{n}}}{1-\mathrm{r}}\right)\\ & \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\mathrm{a}\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)\end{array}$
The above formula does not hold for $r=1$
For $\mathrm{r}=1$, each of the n terms is equal to a, and thus the sum of n terms of the G.P. is $S_n=na$.

The sum of an infinite GP
If a is the first term and $r$ is the common ratio of a G.P. Then,

${\mathrm{S}}_{\mathrm{n}}=\mathrm{a}\left(\frac{1-{\mathrm{r}}^{\mathrm{n}}}{1-\mathrm{r}}\right)=\frac{\mathrm{a}}{1-\mathrm{r}}-\frac{{\mathrm{ar}}^{\mathrm{n}}}{1-\mathrm{r}}$
Let, $-1<r<1$, i.e. $|r|<1$, then

$\underset{n\to \mathrm{\infty }}{lim}{r}^n=0$

[as this is infinite G.P., so $n$ tends to infinity]

$S_{\mathrm{\infty }}=\frac{a}{1-r}$

$S_{\mathrm{\infty }}$ Is the sum of infinite terms of the G.P.

Note: If $r\ge 1$, then the sum of an infinite G.P. tends to infinity.

Application of GP
To write a non-terminating repeating number in p/q form:
Example:
The value of 0.358585858585.

$\begin{array}{rl}& 0.3\overline{58}=0.358585858\dots \dots \text{ to }\mathrm{\infty }\\ & \Rightarrow 0.3+0.058+0.00058+0.0000058+\dots \dots \\ & \Rightarrow \frac{3}{10}+\frac{58}{{10}^3}+\frac{58}{{10}^5}+\frac{58}{{10}^7}+\dots \dots \dots \dots \\ & \Rightarrow \frac{3}{10}+\frac{58}{{10}^3}(1+\frac{1}{{10}^2}+\frac{1}{{10}^4}+\dots .)\\ & \Rightarrow \frac{3}{10}+\frac{58}{{10}^3}\left(\frac{1}{1-\frac{1}{{10}^2}}\right)\\ & \Rightarrow \frac{355}{990}\end{array}$
 

Application of GP
The sum of $n$-term of the series $a+aa+aaa+aaaa+\dots \dots \dots ,\mathrm{\forall }a\in \mathbb{N},1\le a\le 9$ is $\frac{a}{9}[\frac{10}{9}({10}^n-1)-n]$

The sum of $n$-term of the series $0.a+0.aa+0.aaa+0.aaaa+\dots .,\mathrm{\forall }a\in \mathbb{N},1\le a\le 9$ is $\frac{a}{9}\{n-\frac{1}{9}[1-{\left(\frac{1}{10}\right)}^n]\}$