Question

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

Answer 1

nth term of GP = $a{r}^{x-1}$

pth term = $q=a.r^{p-1}$

qth term = $p=a.r^{q-1}$

$\Rightarrow \frac{q}{p}=r^{p-q}$

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

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

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

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

$\Rightarrow P{\left(\frac{q}{p}\right)}^{\frac{p}{p-q}}$

$\Rightarrow \frac{q^{\frac{p}{p-q}}}{p^{p}p-q}-1$

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

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