Key Concepts and Formulas

• Arithmetico-Geometric Progression (AGP): A series where each term is the product of a term from an Arithmetic Progression (AP) and a term from a Geometric Progression (GP). The general form is $a, (a+d)r, (a+2d)r^2, (a+3d)r^3, \dots$.
• Sum to Infinity of an AGP: For an AGP with $|r| < 1$, the sum to infinity ($S_{\infty}$) is given by: $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ where $a$ is the first term of the AP, $d$ is the common difference of the AP, and $r$ is the common ratio of the GP.

Step-by-Step Solution

Step 1: Identify the type of series and its components. The given equation is: $8 = 3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2 p)+\frac{1}{4^3}(3+3 p)+\cdots \cdots \infty$ This is an Arithmetico-Geometric Progression (AGP). We need to identify the first term ($a$), common difference ($d$) of the AP part, and the common ratio ($r$) of the GP part.

• The terms multiplying the powers of $\frac{1}{4}$ form an AP: $3, (3+p), (3+2p), (3+3p), \dots$.

  • The first term of this AP is $a = 3$.
  • The common difference of this AP is $d = (3+p) - 3 = p$.

• The multipliers are powers of $\frac{1}{4}$: $1, \frac{1}{4}, \frac{1}{4^2}, \frac{1}{4^3}, \dots$. This is a GP with the first term $1$ and common ratio $r = \frac{1}{4}$.

Step 2: Check the condition for convergence. The sum to infinity of an AGP exists if the absolute value of the common ratio of the GP is less than 1. Here, $r = \frac{1}{4}$. Since $\left|\frac{1}{4}\right| < 1$, the sum to infinity exists and we can use the formula.

Step 3: Apply the sum to infinity formula for an AGP. The given sum of the series is $S_{\infty} = 8$. Using the formula $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ with $a=3$, $d=p$, and $r=\frac{1}{4}$: $8 = \frac{3}{1 - \frac{1}{4}} + \frac{p \cdot \frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2}$

Step 4: Simplify the terms in the equation. First, calculate the value of $(1-r)$: $1 - \frac{1}{4} = \frac{3}{4}$ Now, substitute this value back into the equation: $8 = \frac{3}{\frac{3}{4}} + \frac{p \cdot \frac{1}{4}}{\left(\frac{3}{4}\right)^2}$ Simplify the first term: $\frac{3}{\frac{3}{4}} = 3 \times \frac{4}{3} = 4$ Simplify the second term: $\frac{p \cdot \frac{1}{4}}{\left(\frac{3}{4}\right)^2} = \frac{\frac{p}{4}}{\frac{9}{16}} = \frac{p}{4} \times \frac{16}{9} = \frac{4p}{9}$ Substitute the simplified terms back into the equation: $8 = 4 + \frac{4p}{9}$

Step 5: Solve for $p$. Subtract 4 from both sides of the equation: $8 - 4 = \frac{4p}{9}$ $4 = \frac{4p}{9}$ Multiply both sides by 9: $4 \times 9 = 4p$ $36 = 4p$ Divide both sides by 4: $p = \frac{36}{4}$ $p = 9$

Common Mistakes & Tips

• Incorrectly identifying $a$ or $d$: Ensure you are picking the correct first term and common difference for the AP part of the series.
• Algebraic errors with fractions: Be meticulous when simplifying fractions, especially when dealing with squares of denominators in the AGP formula.
• Forgetting the convergence condition: Always check if $|r| < 1$ before applying the sum to infinity formula.

Summary

The given infinite series is an Arithmetico-Geometric Progression. By identifying the first term $a=3$, common difference $d=p$, and common ratio $r=\frac{1}{4}$, and applying the formula for the sum to infinity of an AGP, $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$, we set the sum to the given value of 8. Solving the resulting algebraic equation leads to the value of $p=9$.

The final answer is $\boxed{9}$.