Let S Be the Set of All Values of a₁ for Which the Mean Deviation About the Mean of 100 Consecutive Positive Integers a₁, a₂, …, a₁₀₀ Is 25. Then S Is

JEE Main 2023 — 30 January, Evening Shift. Previous Year Question.

Problem

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

Key insight. Shifting a block of 100 consecutive integers up or down doesn’t change how spread out they are around their own mean — only their positions relative to the mean matter for mean deviation, and those positions are identical no matter where the block starts. That single observation is what the whole problem is testing.

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Approach

Rather than trying to guess which values of $a_1$ might work, compute the mean deviation of 100 consecutive integers symbolically. Since $a_i = a_1+i-1$ for $i=1,\ldots,100$, every deviation $|a_i-\bar a|$ depends only on the position $i$, not on $a_1$ itself — so the mean deviation turns out to be a fixed number, completely independent of the starting value $a_1$. Once that’s established, the question becomes: for which $a_1$ is this fixed number equal to 25?

Solution

Step 1 — Write the mean of the 100 integers

For $a_1, a_2, \ldots, a_{100}$ with $a_i=a_1+(i-1)$, the mean is

$$\bar a = a_1+\frac{99}2 = a_1+49.5$$

Step 2 — Express each deviation from the mean

$$a_i-\bar a = \left[a_1+(i-1)\right]-\left[a_1+49.5\right] = (i-1)-49.5$$

Crucially, $a_1$ cancels out completely — the deviation of the $i$-th term from the mean depends only on its position $i$, not on where the sequence starts.

Step 3 — Sum the absolute deviations

For $i=1$ to $100$, $(i-1)-49.5$ runs through $-49.5,-48.5,\ldots,-0.5,0.5,\ldots,49.5$ — symmetric about zero. So the absolute deviations are $0.5, 1.5, \ldots, 49.5$, each occurring twice (once for $i\le50$ and once for $i>50$):

$$\sum_{i=1}^{100}|a_i-\bar a| = 2\left(0.5+1.5+\cdots+49.5\right)$$

The sum $0.5+1.5+\cdots+49.5$ is an AP of 50 terms, first term $0.5$, last term $49.5$:

$$0.5+1.5+\cdots+49.5 = \frac{50}2(0.5+49.5) = 25\times50 = 1250$$

So the total is $2\times1250=2500$.

Step 4 — Compute the mean deviation

$$\text{Mean deviation} = \frac1{100}\sum_{i=1}^{100}|a_i-\bar a| = \frac{2500}{100} = 25$$

Step 5 — Recognize that this holds for every a1

Nothing in this calculation depended on the value of $a_1$ — it canceled out entirely in Step 2. So the mean deviation of any 100 consecutive integers about their own mean is always exactly 25, regardless of where the sequence starts.

Step 6 — Determine the set S

Since the mean deviation equals 25 for every valid starting value, $S$ is simply the set of all $a_1$ for which $a_1, a_2, \ldots, a_{100}$ are all positive integers — that is, every natural number is a valid choice of $a_1$:

$$S = \mathbb N$$

Answer

Common mistakes

• Assuming the problem is asking to solve an equation in $a_1$. Because the question gives a specific numerical mean deviation (25) and asks for the “set of values of $a_1$,” it looks like it should reduce to solving for $a_1$ — but the mean deviation of consecutive integers never actually depends on $a_1$, which is the entire point being tested.
• Trying to verify the result only for a single value of $a_1$, like $a_1=1$. This gives the right numerical mean deviation but doesn’t establish that it holds for every starting value — the symbolic computation in Steps 1–3, where $a_1$ cancels algebraically, is what proves the independence.
• Concluding S is empty or a single value. Since the answer choices for this kind of question typically include options like $\{9\}$, $\varnothing$ or $\{99\}$ alongside $\mathbb N$, it’s tempting to pick a specific finite value without first checking whether the mean deviation genuinely depends on $a_1$ at all.

Practise next

• Find the mean deviation about the mean of 50 consecutive positive integers $a_1,a_2,\ldots,a_{50}$, and confirm — using the same symbolic approach — that it, too, is independent of $a_1$.