1. Key Concepts and Formulas

• Geometric Progression (G.P.): 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 terms can be represented as $a, ar, ar^2, \dots, ar^{n-1}$. The sum of the first $n$ terms is $S_n = \frac{a(r^n-1)}{r-1}$ (for $r \neq 1$). If $a_1, \dots, a_n$ are in G.P. with common ratio $r$, their reciprocals $\frac{1}{a_1}, \dots, \frac{1}{a_n}$ also form a G.P. with common ratio $\frac{1}{r}$.
• Mean ($\bar{X}$): For a set of $N$ numbers $X_1, X_2, \dots, X_N$, the mean is $\bar{X} = \frac{\sum_{i=1}^{N} X_i}{N}$.
• Variance ($\sigma^2$): A measure of the spread of a data set. For $N$ numbers $X_1, \dots, X_N$ and mean $\bar{X}$, the variance is given by $\sigma^2 = \frac{\sum_{i=1}^{N} X_i^2}{N} - (\bar{X})^2$.

2. Step-by-Step Solution

Step 1: Represent the G.P. Terms and Initial Conditions Let the five positive numbers in G.P. be $a_1, a_2, a_3, a_4, a_5$. We can express these terms using the first term $a$ and the common ratio $r$. Since all numbers are positive, we must have $a > 0$ and $r > 0$. The terms are: $a_1 = a$ $a_2 = ar$ $a_3 = ar^2$ $a_4 = ar^3$ $a_5 = ar^4$

Step 2: Formulate Equations from Given Mean Information We are given the mean of the five terms and the mean of their reciprocals.

• Equation from Mean of Terms: The mean of the five terms is $\frac{31}{10}$. $\frac{a + ar + ar^2 + ar^3 + ar^4}{5} = \frac{31}{10}$ Factor out $a$ and multiply by $5$: $a(1 + r + r^2 + r^3 + r^4) = \frac{31}{2}$ The expression in the parenthesis is the sum of a G.P. with first term $1$, common ratio $r$, and $5$ terms. Since the mean of reciprocals is different from $1/(\text{mean of terms})$, we know $r \neq 1$. Using the sum formula $S_n = \frac{a(r^n-1)}{r-1}$: $a \left(\frac{r^5-1}{r-1}\right) = \frac{31}{2} \quad \text{... (1)}$

• Equation from Mean of Reciprocals: The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, \frac{1}{ar^4}$. These also form a G.P. with first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$. The mean of their reciprocals is $\frac{31}{40}$. $\frac{\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4}}{5} = \frac{31}{40}$ Factor out $\frac{1}{a}$ and multiply by $5$: $\frac{1}{a} \left(1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4}\right) = \frac{31}{8}$ The expression in the parenthesis is the sum of a G.P. with first term $1$, common ratio $\frac{1}{r}$, and $5$ terms. $1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4} = \frac{\left(\frac{1}{r}\right)^5 - 1}{\frac{1}{r} - 1} = \frac{\frac{1-r^5}{r^5}}{\frac{1-r}{r}} = \frac{1-r^5}{r^5} \times \frac{r}{1-r} = \frac{-(r^5-1)}{r^4(-(r-1))} = \frac{r^5-1}{r^4(r-1)}$ Substitute this back: $\frac{1}{a} \left(\frac{r^5-1}{r^4(r-1)}\right) = \frac{31}{8} \quad \text{... (2)}$

Step 3: Solve for $a$ and $r$ To solve for $a$ and $r$, we divide Equation (1) by Equation (2): $\frac{a \left(\frac{r^5-1}{r-1}\right)}{\frac{1}{a} \left(\frac{r^5-1}{r^4(r-1)}\right)} = \frac{\frac{31}{2}}{\frac{31}{8}}$ The term $\frac{r^5-1}{r-1}$ cancels from both numerator and denominator on the left side. On the right side, $\frac{31}{2} \div \frac{31}{8} = \frac{31}{2} \times \frac{8}{31} = 4$. $a \cdot (a r^4) = 4$ $a^2 r^4 = 4$ Since $a > 0$ and $r > 0$, we take the square root of both sides: $ar^2 = 2$ This is a significant result: the third term of the G.P., $a_3 = ar^2$, is equal to $2$.

Now, we use the additional given condition: $a_3+a_4+a_5=14$. Substitute the G.P. terms: $ar^2 + ar^3 + ar^4 = 14$ Substitute $ar^2=2$ into this equation: $2 + (ar^2)r + (ar^2)r^2 = 14$ $2 + 2r + 2r^2 = 14$ Divide the entire equation by $2$: $1 + r + r^2 = 7$ Rearrange into a quadratic equation: $r^2 + r - 6 = 0$ Factor the quadratic equation: $(r+3)(r-2) = 0$ This gives two possible values for $r$: $r=-3$ or $r=2$. Since the problem states that the numbers are positive, the common ratio $r$ must also be positive. Therefore, we choose $r=2$.

Now, substitute $r=2$ back into $ar^2=2$ to find $a$: $a(2^2) = 2$ $4a = 2$ $a = \frac{1}{2}$

Step 4: List the G.P. Terms With $a=\frac{1}{2}$ and $r=2$, the five terms of the G.P. are: $a_1 = \frac{1}{2}$ $a_2 = \frac{1}{2} \times 2 = 1$ $a_3 = \frac{1}{2} \times 2^2 = 2$ $a_4 = \frac{1}{2} \times 2^3 = 4$ $a_5 = \frac{1}{2} \times 2^4 = 8$ The G.P. terms are $\left\{\frac{1}{2}, 1, 2, 4, 8\right\}$.

Step 5: Calculate the Variance We need to calculate the variance $\sigma^2 = \frac{m}{n}$ using the terms $\left\{\frac{1}{2}, 1, 2, 4, 8\right\}$ and the mean $\bar{X} = \frac{31}{10}$. We use the formula $\sigma^2 = \frac{\sum X_i^2}{N} - (\bar{X})^2$.

First, calculate the sum of the squares of the terms ($\sum X_i^2$): $\sum X_i^2 = \left(\frac{1}{2}\right)^2 + (1)^2 + (2)^2 + (4)^2 + (8)^2$ $\sum X_i^2 = \frac{1}{4} + 1 + 4 + 16 + 64$ $\sum X_i^2 = \frac{1}{4} + 85 = \frac{1 + 340}{4} = \frac{341}{4}$

Next, calculate the square of the mean $(\bar{X})^2$: $(\bar{X})^2 = \left(\frac{31}{10}\right)^2 = \frac{961}{100}$

Now, substitute these values into the variance formula: $\sigma^2 = \frac{\frac{341}{4}}{5} - \frac{961}{100}$ $\sigma^2 = \frac{341}{20} - \frac{961}{100}$ To subtract these fractions, find a common denominator, which is $100$: $\sigma^2 = \frac{341 \times 5}{20 \times 5} - \frac{961}{100}$ $\sigma^2 = \frac{1705}{100} - \frac{961}{100}$ $\sigma^2 = \frac{1705 - 961}{100}$ $\sigma^2 = \frac{744}{100}$ Simplify the fraction $\frac{744}{100}$ by dividing both the numerator and denominator by their greatest common divisor, which is $4$: $m = 744 \div 4 = 186$ $n = 100 \div 4 = 25$ So, the variance is $\sigma^2 = \frac{186}{25}$.

Step 6: Determine $m+n$ The problem states that the variance is $\frac{m}{n}$, where $m$ and $n$ are co-prime. We have $m=186$ and $n=25$. To verify co-primality: Prime factors of $186$: $2 \times 3 \times 31$ Prime factors of $25$: $5^2$ Since they share no common prime factors, $186$ and $25$ are indeed co-prime. Finally, we calculate $m+n$: $m+n = 186 + 25 = 211$

3. Common Mistakes & Tips

• Ignoring Conditions: Always pay attention to conditions like "positive numbers," as they help eliminate extraneous solutions (e.g., $r=-3$).
• Algebraic Precision: Be meticulous with algebraic manipulations, especially when dealing with sums of G.P. terms and fractions, to avoid calculation errors.
• Choosing Variance Formula: The formula $\sigma^2 = \frac{\sum X_i^2}{N} - (\bar{X})^2$ is generally more efficient for calculations than the definition $\sigma^2 = \frac{\sum (X_i - \bar{X})^2}{N}$, as it avoids multiple subtractions and squaring of decimals or fractions.

4. Summary This problem involved a multi-step approach combining properties of Geometric Progressions with statistical definitions of mean and variance. By representing the G.P. terms and using the given mean of terms and mean of reciprocals, we established a system of equations. Solving these equations led to the determination of the first term $a=\frac{1}{2}$ and common ratio $r=2$, yielding the G.P. terms $\left\{\frac{1}{2}, 1, 2, 4, 8\right\}$. Subsequently, we calculated the variance using the appropriate formula, simplified it to its co-prime form $\frac{186}{25}$, and found $m+n = 211$.

5. Final Answer The final answer is $\boxed{211}$.