If p, q, r are in AP, then prove that pth, qth and rth terms of any GP are in GP.

Solution:

To prove: $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of any GP are in GP.

Given: (i) p, q and r are in AP

The formula used: (i) General term of GP, $T_{n}=a r^{n-1}$

As $p, q, r$ are in A.P.

$\Rightarrow q-p=r-q=d=$ common difference $\ldots$ (i)

Consider a G.P. with the first term as a and common difference $\mathrm{R}$

Then, the $p^{\text {th }}$ term will be $a r^{p-1}$

The $q^{\text {th }}$ term will be $a r^{q-1}$

The $r^{\text {th }}$ term will be $a r^{r-1}$

Considering $p^{\text {th }}$ term and $q^{\text {th }}$ term

$\Rightarrow \frac{q^{\text {th }} \text { term }}{p^{\text {th }} \text { term }}=\frac{\operatorname{ar}^{q-1}}{\operatorname{ar}^{p-1}}$

$\Rightarrow \frac{q^{\text {th }} \text { term }}{p^{\text {th }} \text { term }}=r^{q-1-p+1}$

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

From eqn. (i) $q-p=d$

$\Rightarrow \frac{\mathrm{q}^{\text {th }} \text { term }}{\mathrm{p}^{\text {th }} \text { term }}=\mathrm{r}^{\mathrm{d}}$

Considering $q^{\text {th }}$ term and $r^{\text {th }}$ term

$\Rightarrow \frac{r^{\text {th }} \text { term }}{q^{\text {th }} \text { term }}=\frac{a r^{r-1}}{a r^{q-1}}$

$\Rightarrow \frac{r^{\text {th }} \text { term }}{q^{\text {th }} \text { term }}=r^{r-1-q+1}$

$\Rightarrow \frac{r^{\text {th }} \text { term }}{q^{\text {th }} \text { term }}=r^{r-q}$

From eqn. (i) r – q = d

$\Rightarrow \frac{r^{\text {th }} \text { term }}{q^{\text {th }} \text { term }}=r^{\mathrm{d}}$

We can see that $\mathrm{p}^{\text {th }}, \mathrm{q}^{\text {th }}$ and $\mathrm{r}^{\text {th }}$ terms have common ration i.e $^{r^{d}}$

Hence they are in G.P.

Hence Proved