Let a, b and c be in G.P. with common ratior,

$\mathrm{b}=\mathrm{ar}$

$\mathrm{c}=\mathrm{ar}{ }^{2}$

$3 \mathrm{a}, 7 \mathrm{~b}$ and $15 \mathrm{c}$ are in A.P.

$\Rightarrow 14 b=3 a+15 c$

$\Rightarrow 14(a r)=3 a+15 a r^{2}$

$\Rightarrow 14 r=3+15 r^{2}$

$\Rightarrow 15 r^{2}-14 r+3=0 \quad \Rightarrow(3 r-1)(5 r-3)=0$

$r=\frac{1}{3}, \frac{3}{5}$

Only acceptable value is $\mathrm{r}=\frac{1}{3}$, because

$r \in\left(0, \frac{1}{2}\right]$

$\therefore \mathrm{c} . \mathrm{d}=7 \mathrm{~b}-3 \mathrm{a}=7 \mathrm{ar}-3 \mathrm{a}=\frac{7}{3} \mathrm{a}-3 \mathrm{a}=-\frac{2}{3} \mathrm{a}$

$\therefore 4^{\text {th }}$ term $=15 \mathrm{c}-\frac{2}{3} \mathrm{a}=\frac{15}{9} \mathrm{a}-\frac{2}{3} \mathrm{a}=\mathrm{a}$