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

Hint: 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.