$\because$ Sum of infinite geometric series,
$S_{\infty}=\frac{a}{1-r}$
$\Rightarrow 4=\frac{a}{1-r} \Rightarrow a=4-4 r$...(i)
Also, $T_{2}=a r \Rightarrow a r=\frac{3}{4}$...(ii)
$\therefore $ From Eqs. (i) and (ii), we get
$a=4-4\left(\frac{3}{4 a}\right)$
$\Rightarrow a=4-\frac{3}{a}$
$\Rightarrow a^{2}-4 a+3=0$
$\Rightarrow a^{2}-3 a-a+3=0$
$\Rightarrow a(a-3)-1(a-3)=0$
$\Rightarrow a=1,3$
On putting $a=3$ in Eq. (ii), we get
$3 r=\frac{3}{4}$
$\Rightarrow r=\frac{1}{4}$