Key Concepts and Formulas

1. General Term of a G.P.: For a geometric progression with first term $a$ and common ratio $r$, the $k$-th term is $a_k = ar^{k-1}$.
2. Sum of an Infinite G.P.: The sum of an infinite geometric progression with first term $a$ and common ratio $r$ is $S_\infty = \frac{a}{1-r}$, provided $|r| < 1$.
3. Algebraic Identities: The identities $(x+y)^2 = x^2 + 2xy + y^2$ and $(x-y)^2 = (x+y)^2 - 4xy$ are useful for solving systems of equations involving sums and products of variables.

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Step-by-Step Solution

Step 1: Define the G.P.s and the sequence $c_k$. We are given two G.P.s, $\{a_k\}$ and $\{b_k\}$, with first terms $a_1 = 4$ and $b_1 = 4$, and common ratios $r_1$ and $r_2$ respectively. The $k$-th terms are $a_k = 4r_1^{k-1}$ and $b_k = 4r_2^{k-1}$. We are given $r_1 < r_2$. The sequence $\{c_k\}$ is defined as $c_k = a_k + b_k$.

Step 2: Formulate equations using the given values of $c_2$ and $c_3$. We are given $c_2 = 5$ and $c_3 = \frac{13}{4}$. For $k=2$: $c_2 = a_2 + b_2 = 4r_1^{2-1} + 4r_2^{2-1} = 4r_1 + 4r_2$. Since $c_2 = 5$, we have $4r_1 + 4r_2 = 5$, which simplifies to: $r_1 + r_2 = \frac{5}{4} \quad (*)$ For $k=3$: $c_3 = a_3 + b_3 = 4r_1^{3-1} + 4r_2^{3-1} = 4r_1^2 + 4r_2^2$. Since $c_3 = \frac{13}{4}$, we have $4r_1^2 + 4r_2^2 = \frac{13}{4}$, which simplifies to: $r_1^2 + r_2^2 = \frac{13}{16} \quad (**)$

Step 3: Solve for the common ratios $r_1$ and $r_2$. We use the identity $(r_1+r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2$. Substituting the values from $(*)$ and $(**)$: $(\frac{5}{4})^2 = \frac{13}{16} + 2r_1r_2$ $\frac{25}{16} = \frac{13}{16} + 2r_1r_2$ $2r_1r_2 = \frac{25}{16} - \frac{13}{16} = \frac{12}{16} = \frac{3}{4}$ So, the product of the ratios is $r_1r_2 = \frac{3}{8}$.

Now we find the difference between the ratios using the identity $(r_2-r_1)^2 = (r_1+r_2)^2 - 4r_1r_2$. $(r_2-r_1)^2 = \left(\frac{5}{4}\right)^2 - 4\left(\frac{3}{8}\right)$ $(r_2-r_1)^2 = \frac{25}{16} - \frac{12}{8} = \frac{25}{16} - \frac{24}{16} = \frac{1}{16}$ Taking the square root, $r_2-r_1 = \pm \frac{1}{4}$. Given $r_1 < r_2$, we have $r_2 - r_1 > 0$. Thus, $r_2 - r_1 = \frac{1}{4}$.

We now have a system of linear equations:

1. $r_1 + r_2 = \frac{5}{4}$
2. $-r_1 + r_2 = \frac{1}{4}$ Adding these two equations gives $2r_2 = \frac{6}{4} = \frac{3}{2}$, so $r_2 = \frac{3}{4}$. Substituting $r_2 = \frac{3}{4}$ into the first equation: $r_1 + \frac{3}{4} = \frac{5}{4}$, which gives $r_1 = \frac{2}{4} = \frac{1}{2}$. Thus, $r_1 = \frac{1}{2}$ and $r_2 = \frac{3}{4}$.

Step 4: Calculate $a_6$ and $b_4$. Using the formula $a_k = 4r_1^{k-1}$: $a_6 = 4r_1^{6-1} = 4r_1^5 = 4\left(\frac{1}{2}\right)^5 = 4\left(\frac{1}{32}\right) = \frac{4}{32} = \frac{1}{8}$. Using the formula $b_k = 4r_2^{k-1}$: $b_4 = 4r_2^{4-1} = 4r_2^3 = 4\left(\frac{3}{4}\right)^3 = 4\left(\frac{27}{64}\right) = \frac{108}{64} = \frac{27}{16}$.

Step 5: Calculate the sum of the infinite series $\sum_{k=1}^\infty c_k$. The sum can be split into two convergent infinite G.P.s since $|r_1| = \frac{1}{2} < 1$ and $|r_2| = \frac{3}{4} < 1$. $\sum_{k=1}^\infty c_k = \sum_{k=1}^\infty a_k + \sum_{k=1}^\infty b_k$. The sum of $\{a_k\}$ is $S_{a,\infty} = \frac{a_1}{1-r_1} = \frac{4}{1 - \frac{1}{2}} = \frac{4}{\frac{1}{2}} = 8$. The sum of $\{b_k\}$ is $S_{b,\infty} = \frac{b_1}{1-r_2} = \frac{4}{1 - \frac{3}{4}} = \frac{4}{\frac{1}{4}} = 16$. Therefore, $\sum_{k=1}^\infty c_k = 8 + 16 = 24$.

Step 6: Evaluate the final expression. We need to calculate $\sum_{k=1}^\infty c_k - (12a_6 + 8b_4)$. Substituting the calculated values: $24 - \left(12 \times \frac{1}{8} + 8 \times \frac{27}{16}\right)$ $24 - \left(\frac{12}{8} + \frac{216}{16}\right)$ $24 - \left(\frac{3}{2} + \frac{27}{2}\right)$ $24 - \left(\frac{30}{2}\right)$ $24 - 15 = 9$

Common Mistakes & Tips

• Convergence Check: Always verify that the absolute value of the common ratio is less than 1 before applying the formula for the sum of an infinite G.P.
• Algebraic Accuracy: Pay close attention to fraction arithmetic and algebraic manipulations to avoid errors, especially when solving systems of equations.
• Condition Usage: Ensure that all given conditions, such as $r_1 < r_2$, are used to select the correct roots or values during the solving process.

Summary The problem required finding the common ratios of two geometric progressions using the given information about a combined sequence. We successfully determined $r_1 = \frac{1}{2}$ and $r_2 = \frac{3}{4}$. Subsequently, we calculated specific terms $a_6$ and $b_4$, and the sum of the infinite series $\sum c_k$. Finally, these values were used to evaluate the target expression, yielding a result of 9.

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