Let $a_{1}, a_{2}, \ldots, a_{52}$ be positive integers such that $a_{1}<a_{2}<\ldots<a_{52}$. Suppose, their arithmetic mean is one less than the arithmetic mean of $a_{2}, a_{3}$, $\ldots, a_{52}$. If $a_{52}=100$, then the largest possible value of $a_{1}$ is

We have 52 positive integers arranged in increasing order: $a_1 < a_2 < a_3 < ... < a_{52}$.

The key condition is: The average of all 52 numbers is exactly 1 less than the average of the last 51 numbers.

We call:

• Average of all 52 numbers = $M$
• Average of last 51 numbers = $M + 1$

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From the averages:

• Sum of all 52 numbers = $52M$
• Sum of last 51 numbers = $51(M + 1) = 51M + 51$

Since $a_1 +$ (sum of last 51 numbers) = sum of all 52 numbers:

$a_1 + (51M + 51) = 52M$

$a_1 = 52M - 51M - 51 = M - 51$

This is our key relationship! To maximize $a_1$, we need to maximize $M$.

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To maximize $M$, we need to maximize the sum of all 52 numbers, which means maximizing the sum of the last 51 numbers (since $a_1 = M - 51$).

The constraint: We need $a_1 < a_2 < a_3 < ... < a_{52} = 100$

The strategy: To maximize the sum $a_2 + a_3 + ... + a_{52}$, we should make these numbers as large as possible while maintaining strict inequality.

The optimal arrangement is to make them consecutive integers ending at 100:

• $a_{52} = 100$
• $a_{51} = 99$
• $a_{50} = 98$
• ...
• $a_2 = 50$

So $a_2, a_3, ..., a_{52}$ become: 50, 51, 52, ..., 100

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Sum of consecutive integers from 50 to 100:

Using the formula: Sum = $\frac{\text{(first term + last term)} \times \text{number of terms}}{2}$

Sum = $\frac{(50 + 100) \times 51}{2} = \frac{150 \times 51}{2} = 3825$

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Since sum of last 51 numbers = $51M + 51 = 3825$:

$51M = 3825 - 51 = 3774$

$M = \frac{3774}{51} = 74$

Therefore: $a_1 = M - 51 = 74 - 51 = 23$

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Therefore, the largest possible value of $a_1$ is 23.

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Key Insight: The secret to solving this problem is recognizing that to maximize $a_1$, we need to maximize the sum of the other 51 numbers, which is achieved by making them consecutive integers ending at the given maximum value.