The middle term of an arithmetic series $3,7,11,\cdots 147$ is

$3,7,11,\cdots 147$

It is an arithmetic series

Whose first term $a=3$

Last term $a_n=147$

Common difference $d=7-3=4$

$\Rightarrow a_n=a+(n-1)d$

$\Rightarrow 147=3+(n-1)4$

$\Rightarrow 147-3=(n-1)4$

$\Rightarrow 144=(n-1)4$

$\Rightarrow (n-1)4=144$

$\Rightarrow n-1=36$

$\Rightarrow n=37$

The given series consists of 37 terms

Its middle term will be $\frac{37+1}{2}=$ 19^th term

19^th term $=a+18d$

$=3+18\left(4\right)$

$=3+72$

$=75$.