Question

Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is

• A. 20
• B. 23
• C. 48
• D. 45

Answer 1

The Correct Option is B

Step 1: Problem Understanding

We are given 52 positive integers \( a_1, a_2, ..., a_{52} \), where:

• \( a_1 \) is the least among all
• \( a_{52} = 100 \)
• Average of the 51 values excluding \( a_1 \) is 1 more than the average of all 52 values

We are to find the maximum possible value of \( a_1 \).

Step 2: Let the Average of All 52 Numbers Be \( \bar{A} \)

Then the sum of all 52 values is:

\[ \text{Sum}_{\text{all}} = 52 \bar{A} \]

Average of the other 51 numbers is \( \bar{A} + 1 \), so:

\[ \text{Sum}_{\text{excluding } a_1} = 51(\bar{A} + 1) \]

Subtract to get \( a_1 \):

\[ a_1 = 52 \bar{A} - 51(\bar{A} + 1) = 52\bar{A} - 51\bar{A} - 51 = \bar{A} - 51 \tag{1} \]

Step 3: Assume Arithmetic Progression for \( a_2 \) to \( a_{52} \)

To maximize \( a_1 \), we maximize the sum of the other numbers. The best way is to assume:

• \( a_2, a_3, ..., a_{52} \) form an AP with common difference 1
• So, the average is the middle term: \( a_{27} \)

Also, since \( a_{52} = a_{27} + 25 = 100 \Rightarrow a_{27} = 75 \)

\[ \text{Average of } a_2 \text{ to } a_{52} = 75 \Rightarrow \text{Sum} = 75 \times 51 = 3825 \tag{2} \]

Step 4: Total Sum of All 52 Numbers

From equation (1), average of all 52 numbers is:

\[ \bar{A} = a_1 + 51 \]

So the total sum:

\[ 52\bar{A} = 52(a_1 + 51) = 52a_1 + 2652 \]

But we already know the total must equal \( a_1 + 3825 \)

\[ \Rightarrow a_1 + 3825 = 52a_1 + 2652 \Rightarrow 1173 = 51a_1 \Rightarrow a_1 = \frac{1173}{51} = \boxed{23} \]

Final Answer:

\[ \boxed{a_1 = 23} \]