Question

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

Answer 1

In this question, we are given some notations for G.P. series with n terms. We have to prove that ${{P}^{2}}{{R}^{n}}={{S}^{n}}$ where P is the product of series, R is the sum of reciprocal and S is the sum of series. Here, we will suppose the first term (a) and common ratio (r) of G.P. and use them to find P, R, S in form of a, r and n and then simplify them. Using values obtained we will prove the left-hand side to be equal to the right-hand side. Formula that we will use is given as sum of n terms of $GP=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where, r is common ratio and a is the first term of G.P.

Complete step-by-step solution:
Here, we are given S as the sum of n terms of G.P, P as the product of n terms of GP, and R as the sum of reciprocal of n terms of GP and we have to prove that ${{P}^{2}}{{R}^{n}}={{S}^{n}}$.
Let us first evaluate the values of P, R, and S.
Let us suppose the first term of GP as ’a’ and common ratio as r. Therefore, the series becomes $a,ar,a{{r}^{2}},\cdots \cdots \cdots a{{r}^{n-1}}$.
As we know, sum of n terms of GP is given by $GP=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where, ‘a’ is the first term and r is common ratio. Therefore, for a given series, S is the sum of all terms of GP. Hence,
$S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\cdots \cdots \cdots \cdots \left( 1 \right)$
Let us find P now, since P is product of all n terms, therefore

& P=a\times ar\times a{{r}^{2}}\times a{{r}^{3}}\times \cdots \cdots \cdots a{{r}^{n-1}} \\\ & P=\underbrace{\left( a\times a\times a\times \cdots \cdots \cdots a \right)}_{\text{n times}}\times \left( r\times {{r}^{2}}\times {{r}^{3}}\times \cdots \cdots \cdots {{r}^{n-1}} \right) \\\ \end{aligned}$$ Hence, multiplying 'a' n times, we get $$P={{a}^{n}}\times \left( r\times {{r}^{2}}\times {{r}^{3}}\times \cdots \cdots \cdots {{r}^{n-1}} \right)$$ As we know, ${{x}^{a}}\cdot {{x}^{b}}={{x}^{a+b}}$ using this, we get: $$P={{a}^{n}}\times \left( {{r}^{1+2+3+......+\left( n-1 \right)}} \right)$$ As we know, sum of 1, 2, 3 . . . . . . . n is given as $$1+2+3+.......+n=\dfrac{n\left( n+1 \right)}{2}$$ And for n-1 terms, it becomes $$\dfrac{\left( n-1 \right)\left( n-1+1 \right)}{2}=\dfrac{n\left( n-1 \right)}{2}$$ putting this in above we get: $$P={{a}^{n}}{{r}^{\dfrac{n\left( n-1 \right)}{2}}}\cdots \cdots \cdots \cdots \left( 2 \right)$$ Let us find R, R is the sum of reciprocal of n terms in GP. Therefore, $R=\dfrac{1}{a}+\dfrac{1}{ar}+\dfrac{1}{a{{r}^{2}}}+\cdots \cdots \cdots +$ to n terms. Now since $\dfrac{1}{a}+\dfrac{1}{ar}+\dfrac{1}{a{{r}^{2}}}$ is also a GP with first term as $\dfrac{1}{a}$ and common ratio as $\dfrac{1}{r}$. Also, sum is given as R. As we know, sum of n terms of GP is ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ putting $a=\dfrac{1}{a}\text{ and r}=\dfrac{1}{r}$ we get: $$R=\dfrac{\dfrac{1}{a}\left( {{\left( \dfrac{1}{r} \right)}^{n}}-1 \right)}{\dfrac{1}{r}-1}$$ Taking LCM in denominator, we get: $$R=\dfrac{\dfrac{1}{a}\left( {{\left( \dfrac{1}{r} \right)}^{n}}-1 \right)}{\dfrac{1-r}{r}}$$ Simplifying we get: $$\begin{aligned} & R=\dfrac{1}{a}\left( {{\left( \dfrac{1}{r} \right)}^{n}}-1 \right)\times \dfrac{r}{1-r} \\\ & \Rightarrow \dfrac{1}{a}\left( \dfrac{r}{1-r} \right)\left( \dfrac{1}{{{r}^{n}}}-1 \right) \\\ & \Rightarrow \dfrac{1}{a}\left( \dfrac{r}{1-r} \right)\left( \dfrac{1-{{r}^{n}}}{{{r}^{n}}} \right) \\\ & \Rightarrow \dfrac{r\left( 1-{{r}^{n}} \right)}{a\left( 1-r \right){{r}^{n}}}\cdots \cdots \cdots \cdots \left( 3 \right) \\\ \end{aligned}$$ Now we need to prove ${{P}^{2}}{{R}^{n}}={{S}^{n}}$. Taking left hand side, ${{P}^{2}}{{R}^{n}}$ putting values from (2) and (3) we get: $${{P}^{2}}{{R}^{n}}={{\left( {{a}^{n}}\cdot {{r}^{\dfrac{n\left( n-1 \right)}{2}}} \right)}^{2}}{{\left( \dfrac{r\left( 1-{{r}^{n}} \right)}{a\left( 1-r \right){{r}^{n}}} \right)}^{n}}$$ We know, ${{\left( {{a}^{n}} \right)}^{m}}={{a}^{nm}}$ applying we get: $$\begin{aligned} & {{P}^{2}}{{R}^{n}}={{\left( {{a}^{2n}}{{r}^{n\left( n-1 \right)}} \right)}^{2}}\left( \dfrac{{{\left( 1-{{r}^{n}} \right)}^{n}}}{{{a}^{n}}{{\left( 1-r \right)}^{n}}{{r}^{n\left( n-1 \right)}}} \right) \\\ & \Rightarrow \dfrac{{{a}^{2n}}{{r}^{n\left( n-1 \right)}}{{\left( 1-{{r}^{n}} \right)}^{n}}}{{{a}^{n}}{{\left( 1-r \right)}^{n}}{{r}^{n\left( n-1 \right)}}} \\\ & \Rightarrow \dfrac{{{a}^{n}}\cdot {{a}^{n}}{{r}^{n\left( n-1 \right)}}{{\left( 1-{{r}^{n}} \right)}^{n}}}{{{a}^{n}}{{\left( 1-r \right)}^{n}}{{r}^{n\left( n-1 \right)}}} \\\ \end{aligned}$$ Eliminating ${{a}^{n}}{{r}^{n\left( n-1 \right)}}$ from numerator and denominator we get: $$\Rightarrow \dfrac{{{a}^{n}}{{\left( 1-{{r}^{n}} \right)}^{n}}}{{{\left( 1-r \right)}^{n}}}$$. Since, all terms have power as n, taking n power common from all we get: $$\Rightarrow {{\left( \dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)} \right)}^{n}}$$ From (1) we can see that $\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}=S$. Hence, putting in above we get: ${{S}^{n}}$ which is equal to the right-hand side. **Hence, ${{P}^{2}}{{R}^{n}}={{S}^{n}}$.** **Note:** In this question, students should do all the steps carefully as calculations are complex and there are high chances of making mistakes in taking power. Students can make mistake by taking series as $a,ar,a{{r}^{2}},\cdots \cdots \cdots a{{r}^{n}}$ instead of $a,ar,a{{r}^{2}},\cdots \cdots \cdots a{{r}^{n-1}}$ while taking sum of reciprocal terms, make sure that, common ratio is also changed from r to $\dfrac{1}{r}$.