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If $\text{S}$ is the sum to infinity of a $\text{G.P.}$ whose first term is $'a'$, then the sum of the first $n$ terms is

- $\text{S}\left(1-\dfrac{a}{\text{S}}\right)^n$
- $\text{S}\left[1-\left(1-\dfrac{a}{\text{S}}\right)^n\right]$
- $a\left[1-\left(1-\dfrac{a}{\text{S}}\right)^n\right]$
- $\text{S}\left[1-\left(1-\dfrac{\text{S}}{a}\right)^n\right]$

2 votes

Given that first term = $a$.

$\text{S}=$ sum to infinity of a $\text{G.P} =\dfrac{a}{(1-r)}$

$\Rightarrow r=1-\dfrac{a}{\text{S}}$

$\Rightarrow \text{S}_n=\dfrac{a(1-r^n)}{(1-r)}$

$\Rightarrow \text{S}_n=\text{S}(1-r^n)$

$\Rightarrow \text{S}_n=\text{S}\left[1-\left(1-\dfrac{a}{\text{S}}\right)^n\right]$

$\text{S}=$ sum to infinity of a $\text{G.P} =\dfrac{a}{(1-r)}$

$\Rightarrow r=1-\dfrac{a}{\text{S}}$

$\Rightarrow \text{S}_n=\dfrac{a(1-r^n)}{(1-r)}$

$\Rightarrow \text{S}_n=\text{S}(1-r^n)$

$\Rightarrow \text{S}_n=\text{S}\left[1-\left(1-\dfrac{a}{\text{S}}\right)^n\right]$