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Question: If the sum of first \[n\] terms of the G.P. is \[P\] and sum of their inverse is \[R\], then \[{P^2}...

If the sum of first nn terms of the G.P. is PP and sum of their inverse is RR, then P2{P^2} is equal to
A) RS\dfrac{R}{S}
B) SR\dfrac{S}{R}
C) (RS)n{\left( {\dfrac{R}{S}} \right)^n}
D) (SR)n{\left( {\dfrac{S}{R}} \right)^n}

Explanation

Solution

In this question, we have to find the term which is equal to P2{P^2} in geometric progression. The given problem is the relation of PP and RR. By using the given formula of sum of G.P., we will find the value of PP. Using the formula of the same we will find the value of RR. Then, we will find the relation of PP and RR by using these two relations.

Formula used: We know that,
Let us consider the G.P. be a,ar,ar2,ar3,...,arn−1a,ar,a{r^2},a{r^3},...,a{r^{n - 1}}.
Then, the sum of the series S=a(rn−1)r−1S = \dfrac{{a({r^n} - 1)}}{{r - 1}}

Complete step-by-step answer:
It is given that; the sum of first nn terms of the G.P. is PP and the sum of their inverse is RR.
We have to find the value of P2{P^2}.
Let us consider the G.P. be a,ar,ar2,ar3,...,arn−1a,ar,a{r^2},a{r^3},...,a{r^{n - 1}}.
Then, the sum of the series S=a(rn−1)r−1S = \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, \RightarrowP = {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, \RightarrowR = \dfrac{{\dfrac{1}{a}\left[ {1 - {{\left( {\dfrac{1}{r}} \right)}^n}} \right]}}{{1 - \dfrac{1}{r}}}$$ Simplifying again we get, $\RightarrowR = \dfrac{{r({r^n} - 1)}}{{a{r^n}(r - 1)}} So, we have, $\Rightarrow$$$\dfrac{R}{S} = \dfrac{{r({r^n} - 1)}}{{a{r^n}(r - 1)}} \times \dfrac{{r - 1}}{{a({r^n} - 1)}}
Again,
⇒\Rightarrow$$\dfrac{S}{R} = {({a^2}{r^{n - 1}})^n}From(1)weget, From (1) we get, \dfrac{S}{R} = {P^2}Hence, Hence,\dfrac{S}{R} = {P^2}$$

∴\therefore The correct option is B) SR\dfrac{S}{R}.

Note: In this problem, students can be confused with the calculation of proving the P2{P^2} is equal to SR\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.