Question

Let \[{\rm{S}}\] be the sum, \[{\rm{P}}\] be the product and \[{\rm{R}}\] be the sum of the reciprocals of 3 terms of a G.P. Then \[{P^2}{R^3}:{S^3}\] is equal to
A. \[1:1\]
B. Common ratio: 1
C. \[{({\rm{ first term }})^2}:{({\rm{ common ratio }})^2}\]
D. \[{({\rm{ common ratio }})^n}:1\]

Answer 1

Hint: The three terms of a G.P will initially be assumed to be any three variables. Then, we will add up these three terms to find their sum and compare it to the specified sum. Similar to how we found the provided product, we will also find the product of these three terms. Next, we will add up the inverses of these three G.P. terms and compare it to the supplied sum. We will use these equations to calculate the three generated equations and get the sum of the inverses of the terms and the product of the square of the product of terms.

Formula Used: Product of G.P can be determined using
\[\dfrac{a}{r},a,ar\]
Here, ‘r’ is common ratio

Complete step by step solution: We have been provided in the question that
Let \[{\rm{S}}\] be the sum, \[{\rm{P}}\] be the product and \[{\rm{R}}\] be the sum of the reciprocals of 3 terms of a G.P.
And we are to determine the correct condition for \[{P^2}{R^3}:{S^3}\]
Now, let us consider the three terms of G.P as
\[\dfrac{a}{r},a,ar\]
Now, we have to determine the sum of the terms of a G.P, we have
\[S = \dfrac{a}{r} + a + ar\]
Now, we have to multiply the denominator with the other terms by taking LCM
Sum of G.P is,
\[S = \dfrac{{a({r^2} + r + 1)}}{r}\]
Now, we have to determine the product of the terms of a G.P we have
\[P = \dfrac{a}{r} \times a \times ar\]
On canceling the similar terms, we get
\[P = a \times a \times a\]
That is, the product of the G.P is
\[P = {a^3}\]
Now, we have to determine the sum of the reciprocals of a G.P we have
\[R = \dfrac{r}{a} \times \dfrac{1}{a} \times \dfrac{1}{{ar}}\]
On simplifying the above, we get
\[R = \dfrac{1}{a}\left( {\dfrac{{{r^2} + r + 1}}{r}} \right)\]
Now, let us substitute all the values in the given condition we get
\[\dfrac{{{P^2}{R^3}}}{{{S^3}}} = \dfrac{{{a^6} \cdot \dfrac{1}{{{a^3}}}\left( {\dfrac{{{r^2} + r + 1}}{r}} \right)}}{{{a^3}{{\left( {\dfrac{{{r^2} + r + 1}}{r}} \right)}^3}}}\]
On simplifying, similar terms get cancelled and thus we obtain
\[\dfrac{{{P^2}{R^3}}}{{{S^3}}} = 1\]
Therefore, \[{P^2}{R^3}:{S^3}\] is equal to \[1:1\]

Option ‘A’ is correct

Note: In order to simplify the computation when finding the sum and product of three G.P. terms, we typically assume that the terms are \[\dfrac{a}{r},a,ar\] rather than \[a,ar,a{r^2}\]and this is because \[\dfrac{a}{r},a,ar\] produce \[{a^3}\] when we take their product.