Key Concepts and Formulas

• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The $n$-th term is given by $A_n = A_1 r^{n-1}$.
• Properties of GP Terms: Any term $A_n$ can be expressed relative to another term $A_m$ as $A_n = A_m r^{n-m}$. A useful property for products is that for terms symmetric around a central index, their product can be simplified. For example, $A_a A_b = A_k^2$ if $a+b=2k$.
• Increasing GP of Positive Real Numbers: This implies the first term $A_1 > 0$ and the common ratio $r > 1$.

Step-by-Step Solution

We are given an increasing geometric progression $A_1, A_2, A_3, \dots$ of positive real numbers. We have the following conditions:

1. $A_1 A_3 A_5 A_7 = \frac{1}{1296}$
2. $A_2 + A_4 = \frac{7}{36}$ We need to find the value of $A_6 + A_8 + A_{10}$.

Step 1: Determine the value of $A_4$ using the product condition.

Why this step? The product of terms in a GP often simplifies when expressed in relation to a central term. The indices $1, 3, 5, 7$ are symmetric around the index 4 (since $(1+7)/2 = 4$ and $(3+5)/2 = 4$). This suggests using $A_4$ as the central term.

Let's express each term in the product $A_1 A_3 A_5 A_7$ in terms of $A_4$ and the common ratio $r$:

• $A_1 = A_4 r^{1-4} = A_4 r^{-3}$
• $A_3 = A_4 r^{3-4} = A_4 r^{-1}$
• $A_5 = A_4 r^{5-4} = A_4 r^{1}$
• $A_7 = A_4 r^{7-4} = A_4 r^{3}$

Substitute these into the given product equation: $\left(A_4 r^{-3}\right) \left(A_4 r^{-1}\right) \left(A_4 r^{1}\right) \left(A_4 r^{3}\right) = \frac{1}{1296}$ $A_4^4 \cdot r^{-3-1+1+3} = \frac{1}{1296}$ $A_4^4 \cdot r^0 = \frac{1}{1296}$ $A_4^4 = \frac{1}{1296}$ Since the GP consists of positive real numbers, $A_4$ must be positive. Taking the positive fourth root: $A_4 = \sqrt[4]{\frac{1}{1296}}$ We know that $6^4 = 1296$. $A_4 = \frac{1}{6}$

Step 2: Determine the common ratio $r$ using the sum condition.

Why this step? Now that we have the value of $A_4$, we can use the second condition involving a sum to find the value of another term ($A_2$) and subsequently the common ratio $r$. The common ratio is crucial for calculating any other term in the GP.

The second condition is $A_2 + A_4 = \frac{7}{36}$. Substitute $A_4 = \frac{1}{6}$: $A_2 + \frac{1}{6} = \frac{7}{36}$ Solve for $A_2$: $A_2 = \frac{7}{36} - \frac{1}{6}$ To subtract the fractions, find a common denominator, which is 36: $A_2 = \frac{7}{36} - \frac{6}{36}$ $A_2 = \frac{1}{36}$ We know that $A_4 = A_2 r^{4-2} = A_2 r^2$. We can now find $r^2$: $r^2 = \frac{A_4}{A_2} = \frac{1/6}{1/36} = \frac{1}{6} \times 36 = 6$ Since the progression is an increasing GP of positive real numbers, the common ratio $r$ must be greater than 1. Therefore, we take the positive square root: $r = \sqrt{6}$

Step 3: Calculate the required sum $A_6 + A_8 + A_{10}$.

Why this step? With the values of $A_4$ and $r$ (or $r^2$) known, we can now calculate the terms $A_6, A_8,$ and $A_{10}$ and sum them up. Using $r^2$ can simplify calculations when terms are separated by an even index difference.

We need to find $A_6 + A_8 + A_{10}$. Let's express each term in relation to $A_4$:

• $A_6 = A_4 r^{6-4} = A_4 r^2$
• $A_8 = A_4 r^{8-4} = A_4 r^4 = A_4 (r^2)^2$
• $A_{10} = A_4 r^{10-4} = A_4 r^6 = A_4 (r^2)^3$

Substitute the known values $A_4 = \frac{1}{6}$ and $r^2 = 6$:

• $A_6 = \left(\frac{1}{6}\right) (6) = 1$
• $A_8 = \left(\frac{1}{6}\right) (6)^2 = \left(\frac{1}{6}\right) (36) = 6$
• $A_{10} = \left(\frac{1}{6}\right) (6)^3 = \left(\frac{1}{6}\right) (216) = 36$

Finally, compute the sum: $A_6 + A_8 + A_{10} = 1 + 6 + 36 = 43$

Common Mistakes & Tips

• Sign of the Common Ratio: When solving $r^2 = 6$, remember that an "increasing geometric progression of positive real numbers" requires $r > 1$. Thus, $r = -\sqrt{6}$ is not a valid solution.
• Central Term Strategy: For products of terms in a GP, always look for symmetry in indices to utilize a central term. This often significantly simplifies the problem.
• Efficient Calculation: Expressing terms relative to a known term (like $A_4$) and using powers of $r^2$ when applicable can streamline calculations.

Summary

The problem was solved by first leveraging the product condition to efficiently find the value of the central term $A_4$. Subsequently, the sum condition was used to determine the common ratio $r$. Finally, the required terms $A_6, A_8,$ and $A_{10}$ were calculated using the established values of $A_4$ and $r$, and their sum was computed. The conditions of the problem, such as "increasing" and "positive real numbers," were crucial for selecting the correct value of the common ratio.

The final answer is $\boxed{43}$.