Let the A.P. be $a_r=a+(r-1)d$ where $a=a_1$ and $d$ is the common difference.

We are given $\sum_{r=1}^{\infty} \frac{a_r}{2^r}=4.$

1. Substitute the general term of the A.P.

$\sum_{r=1}^{\infty} \frac{a+(r-1)d}{2^r}=4$

Split the series: $a\sum_{r=1}^{\infty} \frac1{2^r}+d\sum_{r=1}^{\infty} \frac{r-1}{2^r}=4.$

2. Evaluate the standard sums

(i) Geometric series

$\sum_{r=1}^{\infty} \frac1{2^r}=1.$

(ii) Series involving $r-1$

We use $\sum_{r=1}^{\infty} r x^r=\frac{x}{(1-x)^2}, \quad |x|<1.$ For $x=\frac12$, $\sum_{r=1}^{\infty} \frac{r}{2^r}=\frac{1/2}{(1-1/2)^2}=\frac{1/2}{(1/2)^2}=2.$ Hence, $\sum_{r=1}^{\infty} \frac{r-1}{2^r}=\sum_{r=1}^{\infty} \frac r{2^r}-\sum_{r=1}^{\infty} \frac1{2^r}=2-1=1.$

3. Use the given condition

So, $a(1)+d(1)=4$ which gives $a+d=4.$

But $a_2=a+d=4.$ Therefore, $4a_2=4\times 4=16.$

4. Final answer

$\boxed{16}$

The derived answer matches the stored correct answer.