# If the pth and qth terms of a G.P.

Question:

If the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)th term is

Solution:

The $n^{\text {th }}$ term of GP is given by $t_{n}=a r^{n-1}$ where $a$ is the first term and $r$ is the common difference

$\mathrm{p}^{\text {th }}$ term is given as $\mathrm{q}$

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

The above equation can be written as

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

$\Rightarrow \mathrm{q}=\frac{\mathrm{ar}^{\mathrm{P}}}{\mathrm{r}}$

On rearranging the above equation we get

$\Rightarrow \frac{a}{r}=\frac{q}{r p} \ldots$ (a)

$q^{\text {th }}$ term is given as $p$

$\Rightarrow \mathrm{t}_{\mathrm{q}}=\mathrm{ar}^{\mathrm{q}-1}$

$\Rightarrow \mathrm{p}=\mathrm{ar}^{\mathrm{q}-1}$

The above equation can be written as

$\Rightarrow \mathrm{p}=\frac{\mathrm{ar}^{\mathrm{q}}}{\mathrm{r}}$

On rearranging the above equation we get

$\Rightarrow \frac{\mathrm{a}}{\mathrm{r}}=\frac{\mathrm{p}}{\mathrm{r} \mathrm{q}} \ldots$(b)

From equation $(a)$ and $(b)$ we have

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

On rearranging we get

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

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

$(p+q)^{\text {th }}$ term is given by

$\Rightarrow \mathrm{t}_{\mathrm{p}+\mathrm{q}}=\mathrm{a}^{\mathrm{r}^{\mathrm{p}+\mathrm{q}-1}}$

$\Rightarrow \mathrm{t}_{\mathrm{p}+\mathrm{q}}=\left(\mathrm{ar}^{\mathrm{p}-1}\right) \mathrm{r}^{\mathrm{q}}$

But $\mathrm{t}_{\mathrm{p}}=\mathrm{ar}^{\mathrm{p}-1}$ and the $\mathrm{p}^{\text {th }}$ term is $\mathrm{q}$

$\Rightarrow \mathrm{t}_{\mathrm{p}+\mathrm{q}}=\mathrm{q} \mathrm{r}^{\mathrm{q}}$

But

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

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

Using laws of exponents we get

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

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

On rearranging

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

Taking LCM and simplifying we get

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

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

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

Hence the proof.