Question

If $p^{th}$ term of a G. P. is $P$ and its $q^{th}$ term is $Q$. Prove that the $n^{th}$ term is ${\left(\frac{p^{n-q}}{Q^{n-p}}\right)}^{\frac{1}{p-q}}$

Answer 1

We have pth term,

$a_p=p$

$\Rightarrow a{r}^{p-1}=p$

$\Rightarrow a=\frac{p}{r^{p-1}}$.....(1)

qth term,

$a_q=q$

$\Rightarrow a{r}^{q-1}=p$

$\Rightarrow a=\frac{q}{r^{q-1}}$.....(2)

From (1) and (2)

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

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

From (1)

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

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

nth term

$a_n=a{r}^{n-1}$

$={\left(\frac{p^{1-q}}{q^{1-p}}\right)}^{\frac{1}{p-q}}{\left[{\left(\frac{p}{q}\right)}^{\frac{1}{p-q}}\right]}^{n-1}$

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