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Question: If S is the sum to infinity of a G.P whose first term is ‘a’, then the sum of the first n terms is ...

If S is the sum to infinity of a G.P whose first term is ‘a’, then the sum of the first n terms is
A. S(1−aS)nS{\left( {1 - \dfrac{a}{S}} \right)^n}
B. S[1−(1−aS)n]S\left[ {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right]
C. a[1−(1−aS)n]a\left[ {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right]
D. S[1−(1−Sa)n]S\left[ {1 - {{\left( {1 - \dfrac{S}{a}} \right)}^n}} \right]

Explanation

Solution

Since we are given the sum to infinity is S and we know the formula of sum to infinity is a1−r\dfrac{a}{{1 - r}}, Where a is the first term term and r is the common ratio and equating both we get the value of r and now using a and r in the formula of sum to n terms, that is Sn=a(1−rn1−r){S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)we get the required answer.

Complete step by step solution:
We are given that the first term of the GP is a
And the sum to infinity is S
In a geometric progression the sum to infinity is given by the formula
Sum to infinity =a1−r = \dfrac{a}{{1 - r}}
Where a is the first term term and r is the common ratio
Hence we are given the sum to infinity is S
⇒S=a1−r\Rightarrow S = \dfrac{a}{{1 - r}}
From this we can get the value of r by rearranging the terms
⇒S(1−r)=a ⇒1−r=aS ⇒1−aS=r  \Rightarrow S\left( {1 - r} \right) = a \\\ \Rightarrow 1 - r = \dfrac{a}{S} \\\ \Rightarrow 1 - \dfrac{a}{S} = r \\\
So now we have the first term to be a and the common ratio is 1−aS=r1 - \dfrac{a}{S} = r
The Sum to n terms in a GP is given by the formula
⇒Sn=a(1−rn1−r)\Rightarrow {S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)
Substituting the known values we get
⇒Sn=a(1−(1−aS)n1−(1−aS)) ⇒Sn=a(1−(1−aS)n1−(S−aS)) ⇒Sn=a(1−(1−aS)nS−S+aS) ⇒Sn=a(1−(1−aS)naS) ⇒Sn=S(1−(1−aS)n)  \Rightarrow {S_n} = a\left( {\dfrac{{1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}}}{{1 - \left( {1 - \dfrac{a}{S}} \right)}}} \right) \\\ \Rightarrow {S_n} = a\left( {\dfrac{{1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}}}{{1 - \left( {\dfrac{{S - a}}{S}} \right)}}} \right) \\\ \Rightarrow {S_n} = a\left( {\dfrac{{1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}}}{{\dfrac{{S - S + a}}{S}}}} \right) \\\ \Rightarrow {S_n} = a\left( {\dfrac{{1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}}}{{\dfrac{a}{S}}}} \right) \\\ \Rightarrow {S_n} = S\left( {1 - {{\left( {1 - \dfrac{a}{S}} \right)}^n}} \right) \\\
Hence now we obtained the sum of first n terms

Therefore the correct answer is option B.

Note :
If in a sequence of terms, each succeeding term is generated by multiplying each preceding term with a constant value, then the sequence is called a geometric progression. (GP), whereas the constant value is called the common ratio.