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# If the $pth$ and $qth$ terms of a G.P are $q$ and $p$ , respectively, then show that $(p + q)th$ term is ${\left( {\dfrac{{{q^p}}}{{{p^q}}}} \right)^{\dfrac{1}{{p - q}}}}$ .

Last updated date: 13th Jul 2024
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Answer
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Hint: As we know that a geometric sequence is a sequence of nonzero numbers where each term after the first is found by multiplying the previous one by a fixed and nonzero number. The general form of geometric progression is $a,ar,a{r^2},a{r^3},a{r^4}...{a^n}$ , where $a$ is the first term, $r$ is the common ratio. We can solve the given question with the formula i.e. the $nth$ term of a GP is given by ${a_n} = a{r^{n - 1}}$ .

Complete step by step solution:
We will solve the above given question by applying the general formula of the geometric sequence.
Now as per the question we have $pth\,term = q$ , so by applying the general formula we can write it as $a{r^{p - 1}} = q$ . We can assume this as the first equation.
Again we have $qth\,term = p$ , it can be written as $a{r^{q - 1}} = p$ . Let us assume this is the second equation.
On dividing equation first by equation second we get:
$\dfrac{{a{r^{p - 1}}}}{{a{r^{q - 1}}}} = \dfrac{q}{p}$ .
Now we know the basic formula of exponent in division with same base i.e. $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ .
So by applying this we can write
${r^{(p - 1 + q - 1)}} = \dfrac{q}{p} \Rightarrow {r^{(p - q)}} = \dfrac{q}{p}$ .
Now we can transfer the exponent i.e. power to the right hand side of the equation but it will get inverted i.e. it can be written as
$r = {\left( {\dfrac{q}{p}} \right)^{\dfrac{1}{{p - q}}}}$ .
We can substitute this value in the equation second and we get;
$a{\left[ {{{\left( {\dfrac{q}{p}} \right)}^{\dfrac{1}{{p - q}}}}} \right]^{(q - 1)}} = p$ .
We will now solve this, we can write
$a\left[ {{{\left( {\dfrac{q}{p}} \right)}^{\dfrac{{q - 1}}{{q - q}}}}} \right] = p$ . we can isolate the term $a$ in the left hand side of the expression i.e.
$a = p \times \dfrac{1}{{{{\left( {\dfrac{q}{p}} \right)}^{\dfrac{{q - 1}}{{p - q}}}}}}$ .
By further solving it can be written as
$a = p{\left( {\dfrac{p}{q}} \right)^{\dfrac{{q - 1}}{{p - q}}}}$ .
According to the question we have to find the value $(p + q)th$ term or ${a_{(p + q)}}$ . The general formula is ${a_n} = a{r^{n - 1}}$ , so by comparing we have $n = p + q$ .
So by applying the formula we can write ${a_{(p + q)}} = a{r^{(p + q - 1)}}$ .
From the above solution we have the value
$a = p{\left( {\dfrac{p}{q}} \right)^{\dfrac{{q - 1}}{{p - q}}}}$ and $r = {\left( {\dfrac{q}{p}} \right)^{\dfrac{1}{{p - q}}}}$ .
So by substituting the values we have
$p{\left( {\dfrac{p}{q}} \right)^{\dfrac{{q - 1}}{{p - q}}}} \times {\left[ {{{\left( {\dfrac{q}{p}} \right)}^{\dfrac{1}{{p - q}}}}} \right]^{(p + q - 1)}}$ .
On simplifying we can write
$p{\left( {\dfrac{p}{q}} \right)^{\dfrac{{q - 1}}{{p - q}}}} \times \left[ {{{\left( {\dfrac{q}{p}} \right)}^{\dfrac{{p + q - 1}}{{p - q}}}}} \right]$ . we can convert the same base of both the terms inside the bracket, we can write
${\left( {\dfrac{p}{q}} \right)^{\dfrac{{q - 1}}{{p - q}}}}$ as ${\left( {\dfrac{q}{p}} \right)^{\dfrac{{ - (q - 1)}}{{p - q}}}}$ .
We know the exponential formula which says that $\dfrac{a}{b}$ can be written as ${\left( {\dfrac{b}{a}} \right)^{ - 1}}$ . It is also called the inverse formula .
So by putting this we can write $p{\left( {\dfrac{q}{p}} \right)^{\dfrac{{ - (q - 1)}}{{p - q}}}} \times \left[ {{{\left( {\dfrac{q}{p}} \right)}^{\dfrac{{p + q - 1}}{{p - q}}}}} \right]$ .
Now on further simplifying we can write
$p{\left( {\dfrac{q}{p}} \right)^{\dfrac{{ - q + 1 + p + q - 1}}{{p - q}}}}$ . It gives us $p{\left( {\dfrac{q}{p}} \right)^{\dfrac{p}{{p - q}}}}$ .
By breaking the bracket we can write
$p \times \dfrac{{{q^{\dfrac{p}{{p - q}}}}}}{{{p^{\dfrac{p}{{p - q}}}}}}$ . we can see that the term $p$ is in the numerator, so to bring it in the denominator which is also the same base it will be subtracted i.e. $- 1$ from the power of $p$ . So we can write $\dfrac{{{q^{\dfrac{p}{{p - q}}}}}}{{{p^{\dfrac{p}{{p - q}} - 1}}}}$ .
By taking the LCM of the powers in the denominator we have
$\dfrac{{{q^{\dfrac{p}{{p - q}}}}}}{{{p^{\dfrac{{p - p + q}}{{p - q}}}}}} \Rightarrow \dfrac{{{q^{\dfrac{p}{{p - q}}}}}}{{{p^{\dfrac{q}{{p - q}}}}}}$ .
We can take the numerator of the powers out in both the numerator and denominator i.e. $\dfrac{{{q^{p \times \left( {\dfrac{1}{{p - q}}} \right)}}}}{{{p^{q \times \left( {\dfrac{1}{{p - q}}} \right)}}}}$ .
We can write the above expression as ${\left( {\dfrac{{{q^p}}}{{{p^q}}}} \right)^{\dfrac{1}{{p - q}}}}$ .
Hence it is proved that that $(p + q)th$ term is
${\left( {\dfrac{{{q^p}}}{{{p^q}}}} \right)^{\dfrac{1}{{p - q}}}}$ .

Note: Before solving this type of question we should have the proper knowledge of geometric sequence or progression and their formulas. In the above question the basic rule of exponential form is also used. We have used the division exponential rule in the above solution i.e. $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ . We should solve this question carefully by applying the rule of exponents as it can make the solution easier. And we should know that when we know the number of terms in a geometric sequence then it is called a finite geometric sequence. The general formula of finite geometric sequence is $\sum\limits_{i = 0}^n {a{r^n}} = \dfrac{{a(1 - {r^n})}}{{1 - r}}$ .