Question

If sum of 3 terms of a G.P is $S$, product is $P$, and sum of reciprocal of its terms is $R$, then the value of ${P^2}{R^3}$ equal to
A.$S$
B.${S^3}$
C.$2{S^2}$
D.$\dfrac{{{S^2}}}{5}$

Answer 1

We will first assume the three terms of a G.P to be any variable. Then we will find the sum of these three terms and equate it with the given sum. Similarly, we will find the product of these three terms and we will equate it with the given product. Then we will find the sum of reciprocals of these three terms of G.P and equate it with the given sum. We will solve the three obtained equations and obtain the product of the square of product of terms and the sum of reciprocals of the terms using these equations.

Complete step-by-step answer:
Let the three terms of a G.P be $\dfrac{a}{r}$, $a$, $ar$.
Here $r$ is the common ratio.
Now, we will find the sum of these three terms of G.P
${\rm{sum}} = \dfrac{a}{r} + a + ar$
It is given that the sum of a G.P is $S$.
Substituting the value of sum in the above equation, we get
$\Rightarrow S = \dfrac{a}{r} + a + ar$ …….. $\left( 1 \right)$
Now, we will find the product of these three terms of G.P.
${\rm{product}} = \dfrac{a}{r} \times a \times ar$
It is given that the sum of a G.P is $P$.
Substituting the value of product in the above equation, we get
$\Rightarrow P = \dfrac{a}{r} \times a \times ar$
On further simplifying the terms, we get
$\Rightarrow P = {a^3}$ …….. $\left( 2 \right)$
Now, we will find the sum of reciprocals of these three terms of G.P
Sum of reciprocals$= \dfrac{r}{a} + \dfrac{1}{a} + \dfrac{1}{{ar}}$
It is given that the sum of a G.P is $R$.
Substituting the value of sum of reciprocals in the above equation, we get
$\Rightarrow R = \dfrac{r}{a} + \dfrac{1}{a} + \dfrac{1}{{ar}}$
On further simplification, we get
$\Rightarrow R = \dfrac{1}{a}\left( {r + 1 + \dfrac{1}{r}} \right)$ …….. $\left( 3 \right)$
We need to find the value of ${P^2}{R^3}$.
So we will substitute the value of $P$ and $R$ in the expression ${P^2}{R^3}$. Therefore, we get
$\Rightarrow {P^2}{R^3} = {\left( {{a^3}} \right)^2}{\left[ {\dfrac{1}{a}\left( {r + 1 + \dfrac{1}{r}} \right)} \right]^3}$
On applying exponents on the bases, we get
$\Rightarrow {P^2}{R^3} = {a^6}\dfrac{1}{{{a^3}}}{\left( {r + 1 + \dfrac{1}{r}} \right)^3}$
On further simplification, we get
$\Rightarrow {P^2}{R^3} = {a^3}{\left( {r + 1 + \dfrac{1}{r}} \right)^3}$
We can also write this equation as
$\Rightarrow {P^2}{R^3} = {\left[ {a\left( {r + 1 + \dfrac{1}{r}} \right)} \right]^3}$
Multiplying the terms inside the bracket, we get
$\Rightarrow {P^2}{R^3} = {\left( {ar + a + \dfrac{a}{r}} \right)^3}$
From equation $\left( 1 \right)$, we have $S = \dfrac{a}{r} + a + ar$
Therefore, the equation after substituting the value becomes;
$\Rightarrow {P^2}{R^3} = {S^3}$
Thus, the correct option is option B.

Note: Here we have obtained the sum, and product of the G.P. Here G.P means the geometric progression. Geometric progression is defined as a sequence such that every element after the first is obtained by multiplying a constant term to the preceding element. When we have to find the sum and product of three terms of the G.P, we generally assume the terms to be $\dfrac{a}{r}$, $a$ and $ar$instead of $a$, $ar$ and $a{r^2}$ to make the calculation easier because when we take the product of $\dfrac{a}{r}$, $a$ and $ar$, this becomes ${a^3}$.