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+a r+a r^2+\ldots+a r^{n-2}+a r^{n-1}
$
Multiply both sides with $r$

$
r S_n=a r+a r^2+a r^3+\ldots+a r^{n-1}+a r^n
$
Subtract (ii) from (i)

$
\begin{aligned}
& 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{aligned}
$
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 $S_n=n a$.

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

$
\lim _{n \rightarrow \infty} r^n=0
$

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

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

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

Note: If $r \geq 1$, 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{aligned}
& 0.3 \overline{58}=0.358585858 \ldots \ldots \text { to } \infty \\
& \Rightarrow 0.3+0.058+0.00058+0.0000058+\ldots \ldots \\
& \Rightarrow \frac{3}{10}+\frac{58}{10^3}+\frac{58}{10^5}+\frac{58}{10^7}+\ldots \ldots \ldots \ldots \\
& \Rightarrow \frac{3}{10}+\frac{58}{10^3}\left(1+\frac{1}{10^2}+\frac{1}{10^4}+\ldots .\right) \\
& \Rightarrow \frac{3}{10}+\frac{58}{10^3}\left(\frac{1}{1-\frac{1}{10^2}}\right) \\
& \Rightarrow \frac{355}{990}
\end{aligned}
$
 

Application of GP
The sum of $n$-term of the series $a+a a+a a a+a a a a+\ldots \ldots \ldots, \forall a \in \mathbb{N}, 1 \leq a \leq 9$ is $\frac{a}{9}\left[\frac{10}{9}\left(10^n-1\right)-n\right]$

The sum of $n$-term of the series $0 . a+0 . a a+0 . a a a+0 . a a a a+\ldots ., \quad \forall a \in \mathbb{N}, \quad 1 \leq a \leq 9$ is $\frac{a}{9}\left\{n-\frac{1}{9}\left[1-\left(\frac{1}{10}\right)^n\right]\right\}$