Key Concepts and Formulas

• Sum of an Infinite Geometric Progression (G.P.): An infinite G.P. with first term $a$ and common ratio $r$ converges to a finite sum if and only if $|r| < 1$. The sum is given by the formula: $S = \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$
• Difference of Cubes Formula: This algebraic identity is useful for factoring expressions of the form $1 - x^3$: $1 - x^3 = (1-x)(1+x+x^2)$

Step-by-Step Solution

Step 1: Set up equations from the given information. We are given two infinite geometric series. The first series is $\sum_{n=0}^{\infty} a r^n = 57$. Using the formula for the sum of an infinite G.P., we get: $\frac{a}{1-r} = 57 \quad \text{.... (i)}$ The second series is $\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$. This can be rewritten as $\sum_{n=0}^{\infty} (a^3) (r^3)^n$. This is an infinite G.P. with the first term $a^3$ and common ratio $r^3$. Applying the sum formula, we get: $\frac{a^3}{1-r^3} = 9747 \quad \text{.... (ii)}$ For both series to converge, we must have $|r| < 1$ and $|r^3| < 1$, which both imply $|r| < 1$.

Step 2: Relate the two equations to eliminate one variable. Our goal is to find $a + 18r$. We have two equations with two unknowns, $a$ and $r$. A strategic approach is to divide equation (ii) by the cube of equation (i). This will cancel out the term involving $a$. Dividing (ii) by (i)$^3$: $\frac{\frac{a^3}{1-r^3}}{\left(\frac{a}{1-r}\right)^3} = \frac{9747}{57^3}$ Simplifying the left side: $\frac{a^3}{1-r^3} \cdot \frac{(1-r)^3}{a^3} = \frac{(1-r)^3}{1-r^3}$ So, we have: $\frac{(1-r)^3}{1-r^3} = \frac{9747}{57 \cdot 57 \cdot 57} = \frac{9747}{185193}$ Let's simplify the fraction $\frac{9747}{185193}$. We can notice that $185193 = 57 \times 3249$ and $9747 = 57 \times 171$. So, $\frac{9747}{185193} = \frac{171}{3249}$. Further simplification: $171 = 9 \times 19$ and $3249 = 9 \times 361$. So, $\frac{171}{3249} = \frac{19}{361}$. Since $361 = 19^2$, we have $\frac{19}{361} = \frac{1}{19}$. Therefore, $\frac{(1-r)^3}{1-r^3} = \frac{1}{19} \quad \text{.... (iii)}$

Step 3: Use the difference of cubes formula to simplify the equation. We can factor the denominator $1-r^3$ using the difference of cubes formula: $1-r^3 = (1-r)(1+r+r^2)$. Substituting this into equation (iii): $\frac{(1-r)^3}{(1-r)(1+r+r^2)} = \frac{1}{19}$ Assuming $r \neq 1$ (which is true for a convergent G.P. with a finite sum), we can cancel out one factor of $(1-r)$ from the numerator and denominator: $\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$

Step 4: Solve the resulting quadratic equation for r. Cross-multiply the equation from Step 3: $19(1-r)^2 = 1(1+r+r^2)$ Expand $(1-r)^2$: $19(1 - 2r + r^2) = 1 + r + r^2$ Distribute the 19: $19 - 38r + 19r^2 = 1 + r + r^2$ Rearrange the terms to form a standard quadratic equation: $19r^2 - r^2 - 38r - r + 19 - 1 = 0$ $18r^2 - 39r + 18 = 0$ Divide the entire equation by 3 to simplify: $6r^2 - 13r + 6 = 0$ Factor the quadratic equation. We look for two numbers that multiply to $6 \times 6 = 36$ and add up to $-13$. These numbers are $-4$ and $-9$. $6r^2 - 4r - 9r + 6 = 0$ $2r(3r - 2) - 3(3r - 2) = 0$ $(2r - 3)(3r - 2) = 0$ This gives two possible values for $r$: $2r - 3 = 0 \implies r = \frac{3}{2}$ $3r - 2 = 0 \implies r = \frac{2}{3}$

Step 5: Validate the value of r based on the convergence condition. For an infinite geometric series to converge, the common ratio $r$ must satisfy $|r| < 1$. We found two possible values for $r$: $\frac{3}{2}$ and $\frac{2}{3}$. Since $\left|\frac{3}{2}\right| = \frac{3}{2} > 1$, this value of $r$ is invalid for a convergent infinite G.P. The value $r = \frac{2}{3}$ satisfies the condition $\left|\frac{2}{3}\right| < 1$. Thus, this is the correct common ratio.

Step 6: Calculate the value of a. Substitute the valid value of $r = \frac{2}{3}$ into equation (i): $\frac{a}{1-r} = 57$ $\frac{a}{1 - \frac{2}{3}} = 57$ $\frac{a}{\frac{1}{3}} = 57$ $a = 57 \times \frac{1}{3}$ $a = 19$

Step 7: Calculate the final expression $a + 18r$. Now that we have the values of $a$ and $r$, we can compute $a + 18r$: $a + 18r = 19 + 18\left(\frac{2}{3}\right)$ $a + 18r = 19 + \frac{18 \times 2}{3}$ $a + 18r = 19 + 6 \times 2$ $a + 18r = 19 + 12$ $a + 18r = 31$

Common Mistakes & Tips

• Convergence Condition: Always remember that an infinite geometric series only converges if the absolute value of the common ratio $|r|$ is strictly less than 1. Discard any calculated ratio that does not meet this criterion.
• Algebraic Accuracy: Pay close attention to algebraic manipulations, especially when dealing with fractions, exponents, and expanding terms. Mistakes in these areas can lead to incorrect final answers.
• Factoring Quadratics: Practice factoring quadratic equations efficiently. If factoring proves difficult, the quadratic formula can be used as an alternative.

Summary

The problem involved finding the sum of two infinite geometric progressions. We set up equations based on the sum formula and then strategically manipulated these equations to solve for the first term $a$ and the common ratio $r$. The crucial step was to use the difference of cubes factorization and to identify and discard the invalid value of $r$ that did not satisfy the convergence condition $|r| < 1$. Finally, we substituted the valid values of $a$ and $r$ to compute the required expression $a + 18r$.

The final answer is \boxed{31}.