Question

The first term and $n$th term of a G.P. are $a$, $b$ respectively and if $P$ is the product of $n$ terms, prove that ${P^2} = {\left( {ab} \right)^n}$.

Answer 1

Here, we will first use the formula to calculate the sum of first $n$ terms of a geometric progression is $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ to find the number of terms. Then we will use formula of the last term of the G.P. using the formula ${a_n} = a{r^{n - 1}}$, where $r$ is the common ratio and $n$ is the number of terms to find the required values.

Complete step by step answer:

We are given that the first term and $n$th term of a G.P. are $a$, $b$ respectively.
Let us assume that $a$ be the first term of G.P. and $r$ be the common ratio of G.P.
We know that a geometric progression is a sequence of numbers in which each is multiplied by the same factor to obtain the next number in the sequence.
We also know that $n$th term of a G.P. is $a{r^{n - 1}}$, where $a$ be the first term of G.P. and $r$ be the common ratio of G.P, so we have
$\Rightarrow b = a{r^{n - 1}}{\text{ ......eq.(1)}}$
Now, $P$ is the product of $n$ terms, where ${a_1}$, ${a_2}$, ${a_3}$, …,${a_n}$ are the terms, we have
$\Rightarrow P = {a_1} \times {a_2} \times {a_3} \times ... \times {a_n}$
Using the formula of $n$th term of a G.P in the above equation, we get

$$\Rightarrow P = a \times ar \times a{r^2} \times ... \times a{r^{n - 1}} \\\ \Rightarrow P = \left( {a \times a \times a \times ... \times a} \right) \times \left( {r \times {r^2} \times ... \times {r^{n - 1}}} \right) \\\$$

Using the above product $a \times a \times a \times ... \times a$ is $n$ times in the above equation, we get
$\Rightarrow P = {a^n}\left( {r \times {r^2} \times ... \times {r^{n - 1}}} \right)$
Using the power rule, ${a^b}{a^c} = {a^{b + c}}$ in the above equation, we get
$\Rightarrow P = {a^n}{r^{1 + 2 + ... + \left( {n - 1} \right)}}$
Using the formula $1 + 2 + 3 + ... + \left( {n - 1} \right) = \dfrac{{\left( {n - 1} \right)n}}{2}$ in the above equation, we get
$\Rightarrow P = {a^n}{r^{\dfrac{{n\left( {n - 1} \right)}}{2}}}$
We need to prove ${P^2}={\left( {ab} \right)^n}$.
Taking the left-hand side in the above equation using the value of $P$, we get

$$\Rightarrow {\left( {{a^n}{r^{\dfrac{{n\left( {n - 1} \right)}}{2}}}} \right)^2} \\\ \Rightarrow {a^{n \times 2}}{r^{\dfrac{{n\left( {n - 1} \right)}}{2} \times 2}} \\\ \Rightarrow {a^{2n}}{r^{n\left( {n - 1} \right)}} \\\$$

Rewriting the powers of the above equation by using ${a^{bc}} = {\left( {{a^b}} \right)^c}$, we get

$$\Rightarrow {\left( {{a^2}{r^{\left( {n - 1} \right)}}} \right)^n} \\\ \Rightarrow {\left( {a \times a \times {r^{\left( {n - 1} \right)}}} \right)^n} \\\ \Rightarrow {\left( {a \times \left( {a{r^{\left( {n - 1} \right)}}} \right)} \right)^n} \\\$$

Using equation (1) in the above equation, we get

$$\Rightarrow {\left( {a \times b} \right)^n} \\\ \Rightarrow {\left( {ab} \right)^n} \\\$$

Hence, ${P^2} = {\left( {ab} \right)^n}$.

Note: While solving this question, be careful when we are not finding the sum of the geometric series using the formula of first $n$ terms of a geometric progression, $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$. This is key point of the question, some students try to solve this equation directly and end up with a long solution, which is time consuming and mostly lead to wrong answer. Since we know $1 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$ and taking $n = n - 1$ in this formula, we get

$$\Rightarrow 1 + 2 + 3 + ... + \left( {n - 1} \right) = \dfrac{{\left( {n - 1} \right)\left( {n - 1 + 1} \right)}}{2} \\\ \Rightarrow 1 + 2 + 3 + ... + \left( {n - 1} \right) = \dfrac{{n\left( {n - 1} \right)}}{2} \\\$$

We will use this value in the problem.