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

Answer 1

Hint: Since we are given the sum to infinity is S and we know the formula of sum to infinity is $\dfrac{a}{{1 - r}}$, 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 ${S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$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 $ = \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
$ \Rightarrow S = \dfrac{a}{{1 - r}}$
From this we can get the value of r by rearranging the terms
$
   \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 $1 - \dfrac{a}{S} = r$
The Sum to n terms in a GP is given by the formula
$ \Rightarrow {S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$
Substituting the known values we get
$
   \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.