Question

If ${\left( {p + q} \right)^{th}}$ term of a geometric progression be m and ${\left( {p - q} \right)^{th}}$ term be n, then the pth term will be:
(A) $\left( {\dfrac{m}{n}} \right)$
(B) $mn$
(C) $\sqrt {mn} $
(D) $0$

Answer 1

Hint: The given problem involves the concepts of geometric progression. We are given the expressions for a few terms of the GP and we have to find the required term of the same. So, we make use of the formula for the general term of a geometric progression ${a_n} = a{r^{\left( {n - 1} \right)}}$ to solve the problem. We must have a good grip over transposition and simplification rules.

Complete answer:
So, we are given that the ${\left( {p + q} \right)^{th}}$ term of geometric progression is m and ${\left( {p - q} \right)^{th}}$ term is n.
So, we have, ${a_{p + q}} = m$ and ${a_{p - q}} = n$.
Using the formula for general term of a geometric progression, we get,
$ \Rightarrow {a_{p + q}} = a{r^{\left( {p + q - 1} \right)}}$
Substituting the value of ${\left( {p + q} \right)^{th}}$ term, we get,
$ \Rightarrow a{r^{\left( {p + q - 1} \right)}} = m - - - - \left( 1 \right)$
Now, for the ${\left( {p - q} \right)^{th}}$ term, we have,
$ \Rightarrow {a_{p - q}} = a{r^{\left( {p - q - 1} \right)}}$
$ \Rightarrow a{r^{\left( {p - q - 1} \right)}} = n - - - - \left( 2 \right)$
Now, dividing the equation $1$ by equation $2$, we get,
$ \Rightarrow \dfrac{{a{r^{\left( {p + q - 1} \right)}}}}{{a{r^{\left( {p - q - 1} \right)}}}} = \dfrac{m}{n}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow \dfrac{{{r^{\left( {p + q - 1} \right)}}}}{{{r^{\left( {p - q - 1} \right)}}}} = \dfrac{m}{n}$
Using the law of exponents $\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}$, we get,
$ \Rightarrow {r^{\left( {p + q - 1} \right) - \left( {p - q - 1} \right)}} = \dfrac{m}{n}$
$ \Rightarrow {r^{p + q - 1 - p + q + 1}} = \dfrac{m}{n}$
Simplifying the expression, we get,
$ \Rightarrow {r^{2q}} = \dfrac{m}{n}$
Taking square root on both sides of the equation, we get,
$ \Rightarrow {r^q} = \sqrt {\dfrac{m}{n}} $
So, we get the value of ${r^q}$ as $\sqrt {\dfrac{m}{n}} $.
Now, from equation $\left( 1 \right)$, we have,
$ \Rightarrow a{r^{\left( {p + q - 1} \right)}} = m$
Using the law of exponents ${a^x} \times {a^y} = {a^{x + y}}$, we get,
$ \Rightarrow a{r^{\left( {p - 1} \right)}} \times {r^q} = m$
Substituting the value of ${r^q}$ in the equation,
$ \Rightarrow a{r^{\left( {p - 1} \right)}} \times \sqrt {\dfrac{m}{n}} = m$
$ \Rightarrow a{r^{\left( {p - 1} \right)}} = \sqrt {mn} $
Now, we know that pth term of the geometric progression will be of the form $a{r^{\left( {p - 1} \right)}}$.
$ \Rightarrow {a_p} = \sqrt {mn} $
So, we get the pth term as $\sqrt {mn} $.
Hence, the option (C) is correct.

Note:
Geometric progression is a series where any two consecutive terms have the same ratio between them. The common ratio of a geometric series can be calculated by dividing any two consecutive terms of the series. Any term of a geometric progression can be calculated if we know the first term and the common ratio of the series as: ${a_n} = a{r^{\left( {n - 1} \right)}}$.