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Question: If \[S\] is the sum to infinity of a G.P. whose first term is, then the sum of the first\[\;n\] term...

If SS is the sum to infinity of a G.P. whose first term is, then the sum of the first  n\;n terms is

A)  S(1−aS)n B)S[1−(1−aS)n] C)a  [1−(1−aS)n] D)S[1−(1−Sa)n]  A)\,\,S{\left( {1 - \dfrac{a}{S}} \right)^n} \\\ B)S\left[ {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right] \\\ C)a\,\,\left[ {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right] \\\ D)S\left[ {1 - {{\left( {1 - \dfrac{S}{a}} \right)}^n}} \right] \\\
Explanation

Solution

Find the common ratio(r) in progression and substitute in sum of n terms formula
First, since we are given the sum to infinity, we have the formula of it, from which we can derive the value of the common ratio which is . Then after finding this, we have to find the sum of the first  n\;n term by substituting it and then we get the sum of first   n\;n terms.

Complete step by step solution:
We are given that
A series in geometric progression, where
SSis the sum to infinity of a G.P.
First term is “a”.
Now, we have to find the sum of the first  n\;n terms are,
For that we have to find a common ratio. Since we are given the sum of infinite terms
So,
S=a1−r 1−r=aS r=1−aS  S = \dfrac{a}{{1 - r}} \\\ 1 - r = \dfrac{a}{S} \\\ r = 1 - \dfrac{a}{S} \\\ S=a1−r 1−r=aS r=1−aS  S = \dfrac{a}{{1 - r}} \\\ 1 - r = \dfrac{a}{S} \\\ r = 1 - \dfrac{a}{S} \\\
Now, we have found the common ratio of the geometric progression.
We can find the required sum now.
The formula.
Sn=a(1−rn)1−r{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}
Now, we substitute the value and get the value.

=a[1−(1−aS)n]1−(1−aS) =a[1−(1−aS)n]aS   = \dfrac{{a\left[ {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right]}}{{1 - \left( {1 - \dfrac{a}{S}} \right)}} \\\ = \dfrac{{a\left[ {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right]}}{{\dfrac{a}{S}}} \\\ \\\

Cancel a in numerator and denominator.
Then, we get
Sn=S[1−(1−aS)n]{S_n} = S\left[ {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right]
Which is option(B).
Hence, we have found the sum of the first  n\;n terms for the given series in geometric progression.

Therefore, we can say that option ‘B’ is the correct option.

Note: We have to be careful while identifying and substituting in the formula, and identify the formula of sum of infinite terms and finite terms and analyse what is given in the question, so that we can easily substitute in the formula.