Let the positive numbers a1, a2, a3, a4 and a5 be in a G.P. Let their mean and variance be 31/10 and

Q: Why did we choose $a/r^2$ as the first term instead of $a$? Is it just a trick?

It is not just a trick; it is a strategic choice! By centering the G.P. around $a$, the symmetry allows the sum of the terms to be $a(1/r^2 + 1/r + 1 + r + r^2)$. When you look at the sum of the reciprocals, you get $(1/a)(r^2 + r + 1 + 1/r + 1/r^2)$. The common factor is identical, which allows us to eliminate the entire geometric series expression by simple division. It turns a nightmare of algebra into a clean, elegant solution.

Q: I got $r=2$ and $r=1/2$. Why did we reject $r=1/2$?

Great observation! While both values satisfy the mean equations, the problem provides an additional constraint: $a_3 + a_4 + a_5 = 14$. When you test $r=1/2$, the sum of the last three terms is $2 + 1 + 0.5 = 3.5$, which is not 14. Only $r=2$ satisfies this condition. Always check your constraints—they are the gatekeepers of the correct solution.

Q: Can I use the variance formula $\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}$ instead of the shortcut?

Absolutely! You can calculate each $(a_i - \mu)^2$ individually, but it is prone to arithmetic errors because $\mu = 3.1$. The formula $\sigma^2 = \frac{\sum a_i^2}{N} - \mu^2$ is the 'professional' way to handle this. It keeps the numbers cleaner for longer, reducing the chance of a silly mistake in the middle of your calculation.

Q: How do I know if $m$ and $n$ are co-prime?

After you calculate $\sigma^2 = 186/25$, check for common factors. The prime factorization of $25$ is $5^2$. Since $186$ is not divisible by $5$ (it doesn't end in $0$ or $5$), there are no common factors. Thus, they are co-prime by definition. Always perform this check before adding them!

Q: Is there a faster way to find the sum of squares without calculating each term?

Since there are only five terms, calculating them directly is actually the fastest and safest path. However, if you had a larger G.P., you would use the formula for the sum of squares of a G.P., which is $a^2 \frac{r^{2n}-1}{r^2-1}$. For $n=5$, direct calculation is usually faster and less prone to formula-recall errors.

Question

Let the positive numbers $a_1, a_2, a_3, a_4$ and $a_5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_3 + a_4 + a_5 = 14$, then $m + n$ is equal to ____________.

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Geometric Progression

Welcome, fellow traveler on the JEE journey. Today, we are going to dissect a problem that, at first glance, looks like a daunting mountain of algebraic variables. We have five positive numbers in a Geometric Progression (G.P.), and we are given their mean, the mean of their reciprocals, and a specific sum constraint.

Physics and mathematics are rarely about brute force. They are about finding the hidden symmetry that makes the complexity collapse.

### Phase 1: The Art of Choosing Variables

If you label the five terms as $a, ar, ar^2, ar^3, ar^4$, you are setting yourself up for a long, painful algebraic slog. Instead, let us be clever and center our G.P. around the middle term.

We define our terms as:
$$a_1 = \frac{a}{r^2}, \quad a_2 = \frac{a}{r}, \quad a_3 = a, \quad a_4 = ar, \quad a_5 = ar^2$$

When we calculate the sum, we get a beautiful, symmetric expression:
$$S = a \left( \frac{1}{r^2} + \frac{1}{r} + 1 + r + r^2 ight)$$

When we look at the mean of the reciprocals, the expression becomes:
$$	ext{Mean}_{	ext{recip}} = \frac{1}{5a} \left( r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2} ight)$$

Do you see it? The term in the parentheses is identical! By dividing the mean of the terms by the mean of the reciprocals, we get:
$$\frac{a/5}{1/5a} = \frac{31/10}{31/40}$$

This simplifies to $a^2 = 4$. Since the numbers are positive, we immediately find **$a=2$**. The mountain just became a molehill.

### Phase 2: The Hunt for the Common Ratio

Now that we know $a=2$, we substitute it back into our mean equation. We are left with:
$$r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2} = \frac{31}{4}$$

This looks like a quadratic in disguise. Let us introduce a substitution: $t = r + \frac{1}{r}$. Then $r^2 + \frac{1}{r^2} = t^2 - 2$.

Substituting this into our equation gives us:
$$4t^2 + 4t - 35 = 0$$

Solving this quadratic, we find $t = 5/2$. This leads us to $r + 1/r = 5/2$, which gives us two possible values for $r$: $2$ or $1/2$.

But wait! We have one final gatekeeper: the condition $a_3 + a_4 + a_5 = 14$. If we test $r=2$, we get $2 + 4 + 8 = 14$. It fits perfectly!

If we test $r=1/2$, we get $2 + 1 + 0.5 = 3.5$. The choice is clear. Our G.P. is **$1/2, 1, 2, 4, 8$**.

### Phase 3: The Final Calculation of Variance

We have reached the final act. We need the variance $\sigma^2$. We use the elegant formula:
$$\sigma^2 = \frac{\sum a_i^2}{5} - \mu^2$$

We already know $\mu = 31/10$. Now, we calculate the sum of squares:
$$\sum a_i^2 = \left(\frac{1}{2}ight)^2 + 1^2 + 2^2 + 4^2 + 8^2 = 0.25 + 1 + 4 + 16 + 64 = 85.25 = \frac{341}{4}$$

Plugging this into our variance formula:
$$\sigma^2 = \frac{341/4}{5} - \left(\frac{31}{10}ight)^2 = \frac{341}{20} - \frac{961}{100}$$

Converting to a common denominator of $100$, we get:
$$\sigma^2 = \frac{1705 - 961}{100} = \frac{744}{100} = \frac{186}{25}$$

Here, $m=186$ and $n=25$. Since they are co-prime, our final answer is:
$$m + n = 186 + 25 = \mathbf{211}$$

Enter Numerical Value:

Visualized Solution

Defining the G.P. Terms

Mean of the G.P. Terms

Mean of the Reciprocals

Finding the value of $a$

Solving for the Common Ratio $r$ (Part 1)

Solving for the Common Ratio $r$ (Part 2)

Determining the Unique G.P.

Setting up the Variance

Calculating Sum of Squares

Calculating Variance

Final Result $m + n$

The Sigma Insight: Variance and Standard Deviation

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

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