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Question: If \({\left( {p + q} \right)^{th}}\) term of a geometric progression be m and \({\left( {p - q} \rig...

If (p+q)th{\left( {p + q} \right)^{th}} term of a geometric progression be m and (p−q)th{\left( {p - q} \right)^{th}} term be n, then the pth term will be:
(A) (mn)\left( {\dfrac{m}{n}} \right)
(B) mnmn
(C) mn\sqrt {mn}
(D) 00

Explanation

Solution

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 an=ar(n−1){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 (p+q)th{\left( {p + q} \right)^{th}} term of geometric progression is m and (p−q)th{\left( {p - q} \right)^{th}} term is n.
So, we have, ap+q=m{a_{p + q}} = m and ap−q=n{a_{p - q}} = n.
Using the formula for general term of a geometric progression, we get,
⇒ap+q=ar(p+q−1)\Rightarrow {a_{p + q}} = a{r^{\left( {p + q - 1} \right)}}
Substituting the value of (p+q)th{\left( {p + q} \right)^{th}} term, we get,
⇒ar(p+q−1)=m−−−−(1)\Rightarrow a{r^{\left( {p + q - 1} \right)}} = m - - - - \left( 1 \right)
Now, for the (p−q)th{\left( {p - q} \right)^{th}} term, we have,
⇒ap−q=ar(p−q−1)\Rightarrow {a_{p - q}} = a{r^{\left( {p - q - 1} \right)}}
⇒ar(p−q−1)=n−−−−(2)\Rightarrow a{r^{\left( {p - q - 1} \right)}} = n - - - - \left( 2 \right)
Now, dividing the equation 11 by equation 22, we get,
⇒ar(p+q−1)ar(p−q−1)=mn\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,
⇒r(p+q−1)r(p−q−1)=mn\Rightarrow \dfrac{{{r^{\left( {p + q - 1} \right)}}}}{{{r^{\left( {p - q - 1} \right)}}}} = \dfrac{m}{n}
Using the law of exponents axay=ax−y\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}, we get,
⇒r(p+q−1)−(p−q−1)=mn\Rightarrow {r^{\left( {p + q - 1} \right) - \left( {p - q - 1} \right)}} = \dfrac{m}{n}
⇒rp+q−1−p+q+1=mn\Rightarrow {r^{p + q - 1 - p + q + 1}} = \dfrac{m}{n}
Simplifying the expression, we get,
⇒r2q=mn\Rightarrow {r^{2q}} = \dfrac{m}{n}
Taking square root on both sides of the equation, we get,
⇒rq=mn\Rightarrow {r^q} = \sqrt {\dfrac{m}{n}}
So, we get the value of rq{r^q} as mn\sqrt {\dfrac{m}{n}} .
Now, from equation (1)\left( 1 \right), we have,
⇒ar(p+q−1)=m\Rightarrow a{r^{\left( {p + q - 1} \right)}} = m
Using the law of exponents ax×ay=ax+y{a^x} \times {a^y} = {a^{x + y}}, we get,
⇒ar(p−1)×rq=m\Rightarrow a{r^{\left( {p - 1} \right)}} \times {r^q} = m
Substituting the value of rq{r^q} in the equation,
⇒ar(p−1)×mn=m\Rightarrow a{r^{\left( {p - 1} \right)}} \times \sqrt {\dfrac{m}{n}} = m
⇒ar(p−1)=mn\Rightarrow a{r^{\left( {p - 1} \right)}} = \sqrt {mn}
Now, we know that pth term of the geometric progression will be of the form ar(p−1)a{r^{\left( {p - 1} \right)}}.
⇒ap=mn\Rightarrow {a_p} = \sqrt {mn}
So, we get the pth term as mn\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: an=ar(n−1){a_n} = a{r^{\left( {n - 1} \right)}}.