Question

If the sum of first \[n\] terms of the G.P. is \[P\] and sum of their inverse is \[R\], then \[{P^2}\] is equal to
A) \[\dfrac{R}{S}\]
B) \[\dfrac{S}{R}\]
C) \[{\left( {\dfrac{R}{S}} \right)^n}\]
D) \[{\left( {\dfrac{S}{R}} \right)^n}\]

Answer 1

Hint: In this question, we have to find the term which is equal to \[{P^2}\] in geometric progression. The given problem is the relation of \[P\] and \[R\]. By using the given formula of sum of G.P., we will find the value of \[P\]. Using the formula of the same we will find the value of \[R\]. Then, we will find the relation of \[P\] and \[R\] by using these two relations.

Formula used: We know that,
Let us consider the G.P. be \[a,ar,a{r^2},a{r^3},...,a{r^{n - 1}}\].
Then, the sum of the series \[S = \dfrac{{a({r^n} - 1)}}{{r - 1}}\]

Complete step-by-step answer:
It is given that; the sum of first \[n\] terms of the G.P. is \[P\] and the sum of their inverse is \[R\].
We have to find the value of \[{P^2}\].
Let us consider the G.P. be \[a,ar,a{r^2},a{r^3},...,a{r^{n - 1}}\].
Then, the sum of the series \[S = \dfrac{{a({r^n} - 1)}}{{r - 1}}\]
As per the given information,
$\Rightarrow$\[P = a \times ar \times a{r^2} \times .... \times a{r^{n - 1}}\]
Simplifying we get,
$\Rightarrow$\[P = {a^n} \times {r^{\dfrac{{n(n - 1)}}{2}}}\]
Squaring both sides, we get,
$\Rightarrow$\[{P^2} = {a^{2n}} \times {r^{n(n - 1)}}\]
Simplifying we get,
$\Rightarrow$\[{P^2} = {({a^2} \times {r^{n - 1}})^n}\]… (1)
Now,
$\Rightarrow$\[R = \dfrac{1}{a} + \dfrac{1}{{ar}} + \dfrac{1}{{a{r^2}}} + ... + \dfrac{1}{{a{r^{n - 1}}}}\]
Simplifying we get,
$\Rightarrow$\[R = \dfrac{{\dfrac{1}{a}\left[ {1 - {{\left( {\dfrac{1}{r}} \right)}^n}} \right]}}{{1 - \dfrac{1}{r}}}\]
Simplifying again we get,
$\Rightarrow$\[R = \dfrac{{r({r^n} - 1)}}{{a{r^n}(r - 1)}}\]
So, we have,
$\Rightarrow$\[\dfrac{R}{S} = \dfrac{{r({r^n} - 1)}}{{a{r^n}(r - 1)}} \times \dfrac{{r - 1}}{{a({r^n} - 1)}}\]
Again,
$\Rightarrow$\[\dfrac{S}{R} = {({a^2}{r^{n - 1}})^n}\]
From (1) we get,
\[\dfrac{S}{R} = {P^2}\]
Hence, \[\dfrac{S}{R} = {P^2}\]

$\therefore $ The correct option is B) \[\dfrac{S}{R}\].

Note: In this problem, students can be confused with the calculation of proving the \[{P^2}\] is equal to \[\dfrac{S}{R}\]. A sequence is a set of things that are in order. In a geometric sequence each term is found by multiplying the previous term by constant.