Question

If the $pth$ and $qth$ terms of a $GP$ are $q$ and $p$ respectively, then $(p+q)th$ term is

• A. ${\left(\frac{q^p}{p^q}\right)}^{\frac{1}{p-q}}$
• B. ${\left(\frac{q^q}{p^p}\right)}^{\frac{1}{p-q}}$
• C. ${\left(\frac{p^q}{q^p}\right)}^{\frac{1}{p-q}}$
• D. $Noneofthese$

Answer 1

The correct option is A ${\left(\frac{q^p}{p^q}\right)}^{\frac{1}{p-q}}$

Let first four and common ratio of $G.P$ series be $a$ and $r$ respectively

$\therefore p^{th}term=q\Rightarrow {ar}^{p-1}=q........1$

$q^{th}term=p\Rightarrow {ar}^{q-1}=p..........2$

$1÷2$

$\frac{{ar}^{p-1}}{{ar}^{q-1}}=\frac{p}{q}$

or, ${ar}^{p-1-q+1}=\frac{p}{q}\Rightarrow r={\left(\frac{q}{p}\right)}^{\frac{1}{p-q}}$

Putting value of $r$ in $1$

${ar}^{p-1}=q$

Or, $a{\left(\frac{q}{p}\right)}^{\frac{p-1}{p-q}}=q$

Or, $a=q{\left(\frac{p}{q}\right)}^{\frac{p-1}{p-q}}$

$\therefore {p+q}^{th}term={ar}^{q-1}$

$=q{\left(\frac{p}{q}\right)}^{\frac{p-1}{p-q}}{\left(\frac{q}{p}\right)}^{\frac{p+q-1}{p-q}}$

$=q^{(1-\frac{p-1}{p-q}+\frac{p+q+-1}{p-q})}\times p^{(\frac{p-1}{p-q}-\frac{p+q+-1}{p-q})}$

$=q^{\left(\frac{p-q-p+1+p+q-1}{p-q}\right)}\times p^{\left(\frac{p-1-p-q+1}{p-q}\right)}$

$q^{\frac{p}{p-q}}\times q^{-\frac{q}{p-q}}$

$={\left(\frac{q^p}{p^q}\right)}^{\frac{1}{p-q}}$ Ans