1. Key Concepts and Formulas

• Arithmetico-Geometric Progression (AGP): A series where each term is the product of a term from an Arithmetic Progression (A.P.) and a term from a Geometric Progression (G.P.). The general form of an infinite AGP is $S = A + (A+D)x + (A+2D)x^2 + \ldots$.
• Sum of an Infinite AGP: For an infinite AGP to converge, the common ratio $x$ of the G.P. component must satisfy $|x| < 1$. The sum can be found by the method of differences: $S - xS$.
• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant (the common difference, $d$). The $r$-th term is given by $a_r = a_1 + (r-1)d$.

2. Step-by-Step Solution

Step 1: Define the A.P. and express the given sum. Let the arithmetic progression be $a_1, a_2, a_3, \ldots$ with the first term $a_1$ and common difference $d$. Thus, $a_r = a_1 + (r-1)d$. The given sum is an infinite series: $S = \sum_{r=1}^{\infty} \frac{a_r}{2^r} = \frac{a_1}{2^1} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \ldots$ We are given that $S=4$. This series is an Arithmetico-Geometric Progression (AGP) where the A.P. terms are $a_r$ and the G.P. terms are $(1/2)^r$, with the common ratio of the G.P. component being $x = 1/2$. Since $|1/2| < 1$, the series converges.

Step 2: Apply the method of differences for summing an AGP. Write out the series $S$: $S = \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \frac{a_4}{2^4} + \ldots \quad (*)$ Multiply the series by the common ratio of the G.P. component, which is $1/2$: $\frac{1}{2}S = \frac{a_1}{2^2} + \frac{a_2}{2^3} + \frac{a_3}{2^4} + \frac{a_4}{2^5} + \ldots \quad (**)$ Subtract equation $(**)$ from equation $(*)$: $S - \frac{1}{2}S = \left(\frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \ldots \right) - \left(\frac{a_1}{2^2} + \frac{a_2}{2^3} + \frac{a_3}{2^4} + \ldots \right)$ $\frac{1}{2}S = \frac{a_1}{2} + \left(\frac{a_2}{2^2} - \frac{a_1}{2^2}\right) + \left(\frac{a_3}{2^3} - \frac{a_2}{2^3}\right) + \left(\frac{a_4}{2^4} - \frac{a_3}{2^4}\right) + \ldots$ $\frac{1}{2}S = \frac{a_1}{2} + \frac{a_2-a_1}{2^2} + \frac{a_3-a_2}{2^3} + \frac{a_4-a_3}{2^4} + \ldots$

Step 3: Utilize the properties of the Arithmetic Progression. Since $a_1, a_2, a_3, \ldots$ is an A.P. with common difference $d$, we have $a_2-a_1 = d$, $a_3-a_2 = d$, $a_4-a_3 = d$, and so on. Substitute these into the equation from Step 2: $\frac{1}{2}S = \frac{a_1}{2} + \frac{d}{2^2} + \frac{d}{2^3} + \frac{d}{2^4} + \ldots$ The terms $\frac{d}{2^2} + \frac{d}{2^3} + \frac{d}{2^4} + \ldots$ form an infinite geometric series.

Step 4: Sum the resulting Geometric Progression. The infinite geometric series is $\frac{d}{4} + \frac{d}{8} + \frac{d}{16} + \ldots$. The first term of this G.P. is $A = \frac{d}{4}$. The common ratio of this G.P. is $R = \frac{1}{2}$. The sum of this infinite G.P. is given by the formula $\frac{A}{1-R}$: $\text{Sum of G.P.} = \frac{d/4}{1 - 1/2} = \frac{d/4}{1/2} = \frac{d}{4} \times 2 = \frac{d}{2}$ Substitute this sum back into the equation for $\frac{1}{2}S$: $\frac{1}{2}S = \frac{a_1}{2} + \frac{d}{2}$

Step 5: Use the given value of S to find a relationship between $a_1$ and $d$. We are given that $S=4$. Substitute this value into the equation from Step 4: $\frac{1}{2}(4) = \frac{a_1}{2} + \frac{d}{2}$ $2 = \frac{a_1+d}{2}$ Multiply both sides by 2: $4 = a_1+d$

Step 6: Determine the value of $a_2$ and calculate $4a_2$. In an arithmetic progression, the second term $a_2$ is given by $a_2 = a_1 + d$. From Step 5, we found that $a_1+d = 4$. Therefore, $a_2 = 4$. We are asked to find the value of $4a_2$: $4a_2 = 4 \times 4 = 16$

3. Common Mistakes & Tips

• Incorrectly identifying the AGP components: Ensure the A.P. terms are $a_r$ and the G.P. terms are $(1/2)^r$, leading to a common ratio of $1/2$.
• Algebraic errors during subtraction: Be meticulous when subtracting the shifted series to correctly form the new terms.
• Confusing the first term of the original series with the first term of the derived G.P.: In Step 3, the original series starts with $a_1/2$, but the derived G.P. part starts with $d/4$.

4. Summary

The problem requires summing an infinite Arithmetico-Geometric Progression. By applying the standard method of multiplying the series by the common ratio of the geometric component ($1/2$) and subtracting it from the original series, we simplify the expression. This process reveals that $\frac{1}{2}S = \frac{a_1}{2} + \frac{d}{2}$, where $a_1$ is the first term and $d$ is the common difference of the A.P. Given $S=4$, we deduce that $a_1+d=4$. Since $a_2 = a_1+d$ for an A.P., we have $a_2=4$. Consequently, $4a_2 = 4 \times 4 = 16$.

The final answer is $\boxed{16}$.