Question

Let S be the sum, P be the product and R be the sum of reciprocal of n terms in a G.P. Then ${P^2}{R^n}:{S^n}$ is equal to
A) $1:1$
B) ${\left( {{\text{common ratio}}} \right)^n}:1$
C) ${\left( {{\text{first term}}} \right)^2}:{\left( {{\text{common ratio}}} \right)^5}$
D) None of these

Answer 1

Hint: In this question take a series whose terms are in the geometric progression that is $a,ar,a{r^2},a{r^3}, \ldots \ldots \ldots ,a{r^{n - 1}}$. Find the sum of the series using the direct formula for the sum of n terms of G.P, $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$, then find the product of the terms of this series and then the sum of reciprocal, to prove the required.

Complete step-by-step solution:
Let the n terms of G.P be,
$a,ar,a{r^2},a{r^3}, \ldots \ldots \ldots ,a{r^{n - 1}}$
Where $a$ is the first term,
$r$ is the common ratio.
Now it is given that S is the sum of n terms of a G.P,
$ \Rightarrow S = a + ar + a{r^2} + a{r^3} + \ldots \ldots \ldots + a{r^{n - 1}}$
Now as we know that the sum of n terms of a G.P is given by,
$ \Rightarrow S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$................….. (A)
Now take ${n^{th}}$ power on both sides we have,
$ \Rightarrow {S^n} = {\left( {\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}} \right)^n}$
Simplify the terms,
$ \Rightarrow {S^n} = \dfrac{{{a^n}{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}}$..............….. (1)
Now it is given that P is the product of n terms of a G.P,
$ \Rightarrow P = a \times ar \times a{r^2} \times a{r^3} \times \ldots \ldots \ldots \times a{r^{n - 1}}$
Now as we see that $a$ is multiplied by n times then,
$ \Rightarrow P = {a^n}\left( {1 \times r \times {r^2} \times {r^3} \times \ldots \ldots \ldots \times {r^{n - 1}}} \right)$
Now as we see that base $r$ is the same so powers of $r$ are all added up then,
$ \Rightarrow P = {a^n}\left( {{r^{1 + 2 + 3 + \ldots \ldots + n - 1}}} \right)$
Now as we know the sum of the first natural number,
$1 + 2 + 3 + \ldots \ldots \ldots + n = \dfrac{{n\left( {n + 1} \right)}}{2}$
As the last term of power is $\left( {n - 1} \right)$. Replace $n$ with $\left( {n - 1} \right)$ in the equation,
$ \Rightarrow P = {a^n}\left( {{r^{\dfrac{{\left( {n - 1} \right)\left( {n - 1 + 1} \right)}}{2}}}} \right)$
Simplify the term,
$ \Rightarrow P = {a^n}\left( {{r^{\dfrac{{n\left( {n - 1} \right)}}{2}}}} \right)$
Now square both sides,
$ \Rightarrow {P^2} = {\left( {{a^n}\left( {{r^{\dfrac{{n\left( {n - 1} \right)}}{2}}}} \right)} \right)^2}$
Simplify the terms,
$ \Rightarrow {P^2} = {a^{2n}}{r^{n\left( {n - 1} \right)}}$ …………………..….. (2)
Now it is given that R is the sum of reciprocal of n terms of a G.P,
$ \Rightarrow R = \dfrac{1}{a} + \dfrac{1}{{ar}} + \dfrac{1}{{a{r^2}}} + \ldots \ldots \ldots + \dfrac{1}{{a{r^{n - 1}}}}$
Now from equation A replace $a$ with $\dfrac{1}{a}$ and replace $r$ with $\dfrac{1}{r}$ we have,
$ \Rightarrow R = \dfrac{{\dfrac{1}{a}\left( {{{\left( {\dfrac{1}{r}} \right)}^n} - 1} \right)}}{{\dfrac{1}{r} - 1}}$
Take LCM in both numerator and denominator,
$ \Rightarrow R = \dfrac{{\dfrac{1}{a}\left( {\dfrac{{1 - {r^n}}}{{{r^n}}}} \right)}}{{\dfrac{{1 - r}}{r}}}$
Simplify the terms,
$ \Rightarrow R = \dfrac{{\left( {1 - {r^n}} \right)}}{{a{r^{n - 1}}\left( {1 - r} \right)}}$
Multiply numerator and denominator by -1,
$ \Rightarrow R = \dfrac{{\left( {1 - {r^n}} \right)}}{{a{r^{n - 1}}\left( {1 - r} \right)}} \times \dfrac{{ - 1}}{{ - 1}}$
Simplify the terms,
$ \Rightarrow R = \dfrac{{\left( {{r^n} - 1} \right)}}{{a{r^{n - 1}}\left( {r - 1} \right)}}$
Now take ${n^{th}}$ the power of R we have,
$ \Rightarrow {R^n} = {\left( {\dfrac{{\left( {{r^n} - 1} \right)}}{{a{r^{n - 1}}\left( {r - 1} \right)}}} \right)^n}$
Simplify the terms,
$ \Rightarrow {R^n} = \dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{a^n}{r^{n\left( {n - 1} \right)}}{{\left( {r - 1} \right)}^n}}}$ ….. (3)
Now find the value of ${P^2}{R^n}$ by multiplying the equation (2) and (3),
$ \Rightarrow {P^2}{R^n} = {a^{2n}}{r^{n\left( {n - 1} \right)}} \times \dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{a^n}{r^{n\left( {n - 1} \right)}}{{\left( {r - 1} \right)}^n}}}$
Cancel out the common factors,
$ \Rightarrow {P^2}{R^n} = {a^n} \times \dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}}$
Now, from equation (1),
$ \Rightarrow {P^2}{R^n} = {S^n}$
Divide both sides by ${S^n}$,
$ \Rightarrow \dfrac{{{P^2}{R^n}}}{{{S^n}}} = \dfrac{{{S^n}}}{{{S^n}}}$
Cancel out the common factors,
$\therefore \dfrac{{{P^2}{R^n}}}{{{S^n}}} = 1$
Thus, ${P^2}{R^n}:{S^n}$ is equal to 1.

Hence, option (A) is the correct answer.

Note: A series is said to be in G.P if the common ratio that is the ratio of any two consecutive terms of that series remains constant. It is always advised to remember the direct formula of some frequently used series like G.P and A.P, as it helps save a lot of time.