If p, q, r are in A.P., then $p^{th},q^{th}$ and $r^{th}$ terms of any G.P. are themselves in

p, q and r are in A.P

i.e. $2q=p+r$

General term of G.P is $T(K)=a{r}^{(k-1)}$

$p^{th}$ term is :- $T(p)=a{r}^{(p-1)}...(1)$

$q^{th}$ term is :- $T(q)=a{r}^{(q-1)}...(2)$

$r^{th}$ term is :- $T(r)=a{r}^{(R-1)}...(3)$

To prove that ;- $T(p),T(q),T(r)$ are in G.P

$T(q{)}^2=T(p).T(r)$

${\left[a{r}^{(q-1)}\right]}^2=a{r}^{(p-1)}.a{r}^{(R-1)}$

$a^{2}a{r}^{2(q-1)}=a^2{r}^{(p-1)}{r}^{(R-1)}$

$r^{2(q-1)}=r^{(P+R-2)}$

Taking only powers

$2(q-1)=P+R-2$

$2q-2=P+R-2$

$2q=P+R$

This proves that $p^{th},q^{th}$ and $r^{th}$

term are in G.P