Question

Let S be the sum, P the product and R the sum of reciprocal of n terms in a G.P prove that ${P^2} \times {R^n} = {S^n}$.

Answer 1

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

Complete step-by-step answer:

Let the (n) terms G.P be,
$a,ar,a{r^2},a{r^3},..................,a{r^{n - 1}}$ Where (a) is the first term and (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} + .................. + a{r^{n - 1}}$
Now as we know that the sum of (n) terms of a G.P is
$ \Rightarrow S = a\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}$ ........................................ (A)
Now take nth power on both sides we have,
$ \Rightarrow {S^n} = {\left( {a\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}} \right)^n} = {a^n}\dfrac{{{{\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 .................. \times a{r^{n - 1}}$
Now as we see that (a) is multiplied by (n) times therefore
 $ \Rightarrow P = {a^n}\left( {1 \times r \times {r^2} \times {r^3} \times .................. \times {r^{n - 1}}} \right)$
Now as we see that base (r) is same so powers of (r) are all added up therefore.
$ \Rightarrow P = {a^n}\left( {{r^{1 + 2 + 3 + 4 + ........... + \left( {n - 1} \right)}}} \right)$
Now add and subtract by (n) in power of (r) we have,
$ \Rightarrow P = {a^n}\left( {{r^{1 + 2 + 3 + 4 + ........... + \left( {n - 1} \right) + n - n}}} \right)$
Now as we know that sum of first natural number (i.e. 1 +2 +3 +4 +.................+ n) = $\dfrac{{n\left( {n + 1} \right)}}{2}$ so substitute this value in above equation we have,
$ \Rightarrow P = {a^n}\left( {{r^{\dfrac{{n\left( {n + 1} \right)}}{2} - n}}} \right)$
Now simplify the above equation we have,
$ \Rightarrow P = {a^n}\left( {{r^{\dfrac{{n\left( {n + 1} \right) - 2n}}{2}}}} \right) = {a^n}\left( {{r^{\dfrac{{n\left( {n - 1} \right)}}{2}}}} \right)$
Now squaring on both sides we have,
$ \Rightarrow {P^2} = {\left( {{a^n}\left( {{r^{\dfrac{{n\left( {n - 1} \right)}}{2}}}} \right)} \right)^2} = {a^{2n}}{r^{n\left( {n - 1} \right)}}$......................... (2)
Now it is also given that sum of reciprocal of (n) terms in a G.P is R therefore we have,
$ \Rightarrow R = \dfrac{1}{a} + \dfrac{1}{{ar}} + \dfrac{1}{{a{r^2}}} + .................. + \dfrac{1}{{a{r^{n - 1}}}}$
Now from equation (A) replace (a) with (1/a) and replace (r) with (1/r) we have,
$ \Rightarrow R = \dfrac{1}{a}\dfrac{{\left( {{{\left( {\dfrac{1}{r}} \right)}^n} - 1} \right)}}{{\left( {\dfrac{1}{r} - 1} \right)}}$
$ \Rightarrow R = \dfrac{1}{a}\dfrac{{\left( {1 - {{\left( {\dfrac{1}{r}} \right)}^n}} \right)}}{{\left( {1 - \dfrac{1}{r}} \right)}} = \dfrac{1}{a}\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}\dfrac{r}{{{r^n}}} = \dfrac{1}{a}\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}\dfrac{1}{{{r^{n - 1}}}}$
Now take nth power of R we have,
\[ \Rightarrow {R^n} = {\left( {\dfrac{1}{a}\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}\dfrac{1}{{{r^{n - 1}}}}} \right)^n}\]
Now simplify the above equation we have,
$ \Rightarrow {R^n} = \dfrac{1}{{{a^n}}}\dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}}{\left( {\dfrac{1}{{{r^{n - 1}}}}} \right)^n} = \dfrac{1}{{{a^n}}}\dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}}\dfrac{1}{{{r^{n\left( {n - 1} \right)}}}}$............... (3)
Now multiply equation (2) and (3) together we have,
$ \Rightarrow {P^2} \times {R^n} = {a^{2n}}{r^{n\left( {n - 1} \right)}} \times \dfrac{1}{{{a^n}}}\dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}}\dfrac{1}{{{r^{n\left( {n - 1} \right)}}}}$
Now simplify the above equation we have,
$ \Rightarrow {P^2} \times {R^n} = {a^n} \times \dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}}$
Now from equation (1) we have,
$ \Rightarrow {P^2} \times {R^n} = {a^n} \times \dfrac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}} = {S^n}$
$ \Rightarrow {P^2} \times {R^n} = {S^n}$
Hence Proved.

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.