Let S be the set of all values of a1 for which the mean deviation about the mean of 100 consecutive

Q: Wait, does the mean deviation really not depend on the starting value $a_1$? That feels wrong.

It is a beautiful property of this specific problem! Because mean deviation is a measure of spread (dispersion), and shifting a set of numbers by a constant $a_1$ does not change the relative distances between the numbers, the spread remains identical. When you subtract the mean from each term, the $a_1$ term in the observation and the $a_1$ term in the mean cancel each other out perfectly. You are left with the same set of deviations regardless of where you start.

Q: Why is the mean $\bar{x} = a_1 + 49.5$? How did we get that?

Great question. The mean of an arithmetic progression is simply the average of the first and last terms. Here, the first term is $a_1$ and the 100th term is $a_1 + 99$. So, $\bar{x} = \frac{a_1 + (a_1 + 99)}{2} = \frac{2a_1 + 99}{2} = a_1 + 49.5$. It sits exactly in the middle of the sequence, which is why it's the perfect reference point for the deviations.

Q: Can we use variance instead of mean deviation to solve this?

You absolutely could! Variance is also independent of the shift $a_1$. However, the problem specifically asks for mean deviation, so we must stick to the absolute differences. The symmetry trick we used for the mean deviation is actually very similar to how you would simplify the variance calculation. Both methods highlight the same underlying geometric reality.

Question

Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \dots, a_{100}$ is 25. Then $S$ is

Options:

(A) $\phi$

(B) $\{99\}$

(C) $\mathbb{N}$

(D) $\{9\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, infinite number line. You are tasked with picking 100 consecutive positive integers, starting from some arbitrary integer $a_1$.

You lay them out as $a_1, a_1+1, a_1+2, \dots, a_1+99$. At first glance, this seems like a problem that depends heavily on your choice of $a_1$.

However, as we dive into the mathematics, we will uncover a hidden, elegant truth: the spread of these numbers is entirely independent of where you start.

### The Center of Gravity

To understand the mean deviation, we must first find the mean, $\bar{x}$. The mean is the center of gravity of our data.

For any arithmetic progression, the mean is simply the average of the first and last terms. Our sequence starts at $a_1$ and ends at $a_1 + 99$.

Thus, the mean is:

$$\bar{x} = \frac{a_1 + (a_1 + 99)}{2} = a_1 + 49.5$$

Notice how the mean sits exactly in the middle of our sequence. It is the anchor point from which we measure the deviation of every other number.

### The Magic of Cancellation

Now, let us look at the mean deviation formula: $MD = \frac{1}{100} \sum_{i=1}^{100} |a_i - \bar{x}|$.

When we substitute our general term $a_i = a_1 + i - 1$ and our mean $\bar{x} = a_1 + 49.5$ into this expression, something magical happens:

$$|a_i - \bar{x}| = |(a_1 + i - 1) - (a_1 + 49.5)| = |i - 50.5|$$

The $a_1$ terms vanish! They cancel out completely.

This is the moment of realization: the mean deviation does not care about the starting value $a_1$. Whether you start at 1 or 1,000,000, the "shape" of the deviation remains identical.

### Exploiting Symmetry

We are now left with the sum $\sum_{i=1}^{100} |i - 50.5|$. If we write out the first few terms, we see a pattern: $|1 - 50.5| = 49.5$, $|2 - 50.5| = 48.5$, and so on.

As we approach the middle, the values shrink, and then they grow again on the other side. The sequence of deviations is perfectly symmetric:

$$49.5, 48.5, \dots, 0.5, 0.5, \dots, 48.5, 49.5$$

Instead of summing all 100 terms, we can sum the first 50 and multiply by 2. The sum of the first 50 terms is an arithmetic progression: $0.5 + 1.5 + \dots + 49.5$.

Using the sum formula $S_n = \frac{n}{2}(a + l)$, where $n=50$, $a=0.5$, and $l=49.5$, we get:

$$Sum = 2 	imes \left[ \frac{50}{2} (0.5 + 49.5) ight] = 2 	imes (25 	imes 50) = 2500$$

### The Final Result

Finally, we divide this sum by the total number of observations, 100:

$$MD = \frac{2500}{100} = 25$$

The problem stated that the mean deviation is 25, and we have just proven that it is **always 25** for any set of 100 consecutive integers.

Since $a_1$ can be any positive integer, the set $S$ of all possible values for $a_1$ is the set of all natural numbers, $\mathbb{N}$. We have conquered the problem not by brute force, but by understanding the symmetry of the numbers themselves.

Animated Solution for Mathematics - Statistics: Let $S$ be the set of all values of $a_{1}$ for which the mean deviation about the mean of 100 consecutive positive integers $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ is 25. Then $S$ is

Select Answer:

Visualized Solution

Define the Sequence

Formula for Mean $\overset{x}{ˉ}$

Evaluate the Mean

Setup Mean Deviation

Simplify the Absolute Term

Analyze the Summation Symmetry

Exploit Symmetry

Calculate the Total Sum

Final Result and Conclusion

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Center of Gravity

The Magic of Cancellation

Exploiting Symmetry

The Final Result

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