Let $$a_1, a_2, a_3, \ldots$$ be an A.P. If $$\sum_{r=1}^{\infty} \frac{a_r}{2^r} = 4$$, then $$4a_2$$ is equal to ______

We have an A.P. $$a_1, a_2, a_3, \ldots$$ with first term $$a_1 = a$$ and common difference $$d$$, so $$a_r = a + (r-1)d$$. We are given that $$\sum_{r=1}^{\infty}\frac{a_r}{2^r} = 4$$.

We split this into two sums: $$\sum_{r=1}^{\infty}\frac{a + (r-1)d}{2^r} = a\sum_{r=1}^{\infty}\frac{1}{2^r} + d\sum_{r=1}^{\infty}\frac{r-1}{2^r}$$.

We know $$\sum_{r=1}^{\infty}\frac{1}{2^r} = 1$$. For the second sum, we write $$\sum_{r=1}^{\infty}\frac{r-1}{2^r} = \sum_{r=2}^{\infty}\frac{r-1}{2^r} = \sum_{k=1}^{\infty}\frac{k}{2^{k+1}} = \frac{1}{2}\sum_{k=1}^{\infty}\frac{k}{2^k}$$.

Using the standard result $$\sum_{k=1}^{\infty}kx^k = \frac{x}{(1-x)^2}$$ with $$x = \frac{1}{2}$$, we get $$\sum_{k=1}^{\infty}\frac{k}{2^k} = \frac{1/2}{(1/2)^2} = 2$$. So $$\sum_{r=1}^{\infty}\frac{r-1}{2^r} = \frac{1}{2}\cdot 2 = 1$$.

Therefore $$a\cdot 1 + d\cdot 1 = 4$$, which gives $$a + d = 4$$. Since $$a_2 = a + d$$, we have $$a_2 = 4$$, and thus $$4a_2 = 16$$.

Hence, the correct answer is 16.