GO Classes CS/DA 2025 | Weekly Quiz 1 | Fundamental Course | Question: 5

Answer 1

Given that first term = $a$.

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

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

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

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

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

Comments

put value of r in denominator

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for infinite sum in G.P, S = a/(1-r)

just split the numerator like: Sn =[ a/(1-r) ] - rⁿ [ a/(1-r) ]

then Sn = S - (rⁿ)S

=> Sn = S( 1-rⁿ)

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a/1-r is nothing but the sum