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Question: In a G.P. if \({{\left( m+n \right)}^{th}}\) term is p and \({{\left( m-n \right)}^{th}}\) term is q...

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

Explanation

Solution

Hint: First, we should know the formula which we are going to use is nth{{n}^{th}} term finding formula Tn=arn−1{{T}_{n}}=a{{r}^{n-1}} . Then, we have to consider our given terms as (m+n)th{{\left( m+n \right)}^{th}} term as p and (m−n)th{{\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 mth{{m}^{th}} term in Geometric progression (G.P.) where it is given as
(m+n)th=p{{\left( m+n \right)}^{th}}=p ……………………(1)
(m−n)th=q{{\left( m-n \right)}^{th}}=q …………………..(2)
Now, we know the general form of G.P. is given as:
a,ar,ar2,ar3,....a,ar,a{{r}^{2}},a{{r}^{3}},....
So, to find the nth{{n}^{th}} formula in series of G.P. is given as:
⇒Tn=arn−1\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 nth{{n}^{th}} term which we want to find in series.
So, we can represent our given equation in from of equation (3) as,
T(m+n)=ar(m+n)−1{{T}_{\left( m+n \right)}}=a{{r}^{\left( m+n \right)-1}} ………………………….(4) Here, n is m+nm+n
T(m−n)=ar(m−n)−1{{T}_{\left( m-n \right)}}=a{{r}^{\left( m-n \right)-1}} ………………………….(5) Here, n is m−nm-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=ar(m+n)−1p=a{{r}^{\left( m+n \right)-1}} …………………………….(6)
q=ar(m−n)−1q=a{{r}^{\left( m-n \right)-1}} ……………………………(7)
Now, we will multiply both the above equations. We get as.
⇒pq=a2rm+n−1⋅rm−n−1\Rightarrow pq={{a}^{2}}{{r}^{m+n-1}}\cdot {{r}^{m-n-1}}
Using the multiplication rule i.e. an×am=an+m{{a}^{n}}\times {{a}^{m}}={{a}^{n+m}}
So, on simplification, we get
⇒pq=a2rm+n−1+m−n−1\Rightarrow pq={{a}^{2}}{{r}^{m+n-1+m-n-1}}
⇒pq=a2r2m−2\Rightarrow pq={{a}^{2}}{{r}^{2m-2}} (Cancelling all the positive negative terms)
⇒pq=a2r2(m−1)\Rightarrow pq={{a}^{2}}{{r}^{2\left( m-1 \right)}}
⇒pq=(ar(m−1))2\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
⇒pq=ar(m−1)=Tm\Rightarrow \sqrt{pq}=a{{r}^{\left( m-1 \right)}}={{T}_{m}}
Thus, the required answer of mth{{m}^{th}} term is pq\sqrt{pq}.
Hence, option (a) is the correct answer.

Note: Another approach to solve this kind of problem is as given below:
p=ar(m+n)−1p=a{{r}^{\left( m+n \right)-1}} ……(1)
q=ar(m−n)−1q=a{{r}^{\left( m-n \right)-1}} …….(2)
Now, we will be dividing both the equation,
pq=arm+n−1arm−n−1\dfrac{p}{q}=\dfrac{a{{r}^{m+n-1}}}{a{{r}^{m-n-1}}}
Here using division rule of same coefficient i.e. aman=am−n\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}
∴pq=rm+n−1−(m−n−1)\therefore \dfrac{p}{q}={{r}^{m+n-1-\left( m-n-1 \right)}} . So, on solving we will get as
∴pq=r2n⇒(pq)12n=r\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(qp)m+n−12na=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 mth{{m}^{th}} term i.e. Tm=arm−1{{T}_{m}}=a{{r}^{m-1}} , and on solving we will get the same answer as we got i.e. pq\sqrt{pq} .