Question

In a G.P. if ${{\left( m+n \right)}^{th}}$ term is p and ${{\left( m-n \right)}^{th}}$ term is q, then its ${{m}^{th}}$ term is
(a)$\sqrt{pq}$
(b)$\sqrt{\left( \dfrac{p}{q} \right)}$
(c)$\sqrt{\left( \dfrac{q}{p} \right)}$
(d)$\dfrac{p}{q}$

Answer 1

Hint: First, we should know the formula which we are going to use is ${{n}^{th}}$ term finding formula ${{T}_{n}}=a{{r}^{n-1}}$ . Then, we have to consider our given terms as ${{\left( m+n \right)}^{th}}$ term as p and ${{\left( m-n \right)}^{th}}$ term as q. Then, on getting both equations we will multiply them and on further simplification, we get our answer.

Complete step-by-step answer:
In this question, we are supposed to find the ${{m}^{th}}$ term in Geometric progression (G.P.) where it is given as
${{\left( m+n \right)}^{th}}=p$ ……………………(1)
${{\left( m-n \right)}^{th}}=q$ …………………..(2)
Now, we know the general form of G.P. is given as:
$a,ar,a{{r}^{2}},a{{r}^{3}},....$
So, to find the ${{n}^{th}}$ formula in series of G.P. is given as:
$\Rightarrow {{T}_{n}}=a{{r}^{n-1}}$ ……………………………………….(3)
Where a is the first term in series, r is a common ratio and n is the ${{n}^{th}}$ term which we want to find in series.
So, we can represent our given equation in from of equation (3) as,
${{T}_{\left( m+n \right)}}=a{{r}^{\left( m+n \right)-1}}$ ………………………….(4) Here, n is $m+n$
${{T}_{\left( m-n \right)}}=a{{r}^{\left( m-n \right)-1}}$ ………………………….(5) Here, n is $m-n$
Now, we are assuming equation (4) and (5) as some constant variable let say p and q respectively, as it is given to us.
So, rewriting equation (4) and (5) again, we get
$p=a{{r}^{\left( m+n \right)-1}}$ …………………………….(6)
$q=a{{r}^{\left( m-n \right)-1}}$ ……………………………(7)
Now, we will multiply both the above equations. We get as.
$\Rightarrow pq={{a}^{2}}{{r}^{m+n-1}}\cdot {{r}^{m-n-1}}$
Using the multiplication rule i.e. ${{a}^{n}}\times {{a}^{m}}={{a}^{n+m}}$
So, on simplification, we get
$\Rightarrow pq={{a}^{2}}{{r}^{m+n-1+m-n-1}}$
$\Rightarrow pq={{a}^{2}}{{r}^{2m-2}}$ (Cancelling all the positive negative terms)
$\Rightarrow pq={{a}^{2}}{{r}^{2\left( m-1 \right)}}$
$\Rightarrow pq={{\left( a{{r}^{\left( m-1 \right)}} \right)}^{2}}$
The above equation is in the form of G.P. formula as given in equation (3). So, here m is the term we want to find an answer for. So, we get as
$\Rightarrow \sqrt{pq}=a{{r}^{\left( m-1 \right)}}={{T}_{m}}$
Thus, the required answer of ${{m}^{th}}$ term is $\sqrt{pq}$.
Hence, option (a) is the correct answer.

Note: Another approach to solve this kind of problem is as given below:
$p=a{{r}^{\left( m+n \right)-1}}$ ……(1)
$q=a{{r}^{\left( m-n \right)-1}}$ …….(2)
Now, we will be dividing both the equation,
$\dfrac{p}{q}=\dfrac{a{{r}^{m+n-1}}}{a{{r}^{m-n-1}}}$
Here using division rule of same coefficient i.e. $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
$\therefore \dfrac{p}{q}={{r}^{m+n-1-\left( m-n-1 \right)}}$ . So, on solving we will get as
$\therefore \dfrac{p}{q}={{r}^{2n}}\Rightarrow {{\left( \dfrac{p}{q} \right)}^{\dfrac{1}{2n}}}=r$
So, putting this value or r in equation (1) for finding value of a. So, we will get value of a as:
$a=p{{\left( \dfrac{q}{p} \right)}^{\dfrac{m+n-1}{2n}}}$
Now, we have the value of r and a. So, we will directly put in formula of ${{m}^{th}}$ term i.e. ${{T}_{m}}=a{{r}^{m-1}}$ , and on solving we will get the same answer as we got i.e. $\sqrt{pq}$ .