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

The correct answer is (B):
We want to maximize the value of a1, subject to the condition that a1 is the least of the 52 numbers and that the average of 51 numbers (excluding a1) is 1 less than the average of all the 52 numbers. Since a52 is 100 and all the numbers are positive integers, maximizing a1 entails maximizing a2, a3, ….a51.
The only way to do this is to assume that a2, a3…. a52 are in an AP with a common difference of 1.
Let the average of a2, a3…. a52 i.e. a27 be A.
(Note: The average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)
Since a52 = a27 + 25 and a52 = 100
=> A = 100 – 25 = 75
a2 + a3 + … + a52 = 75×51 = 3825
Given a1 + a2 +… + a52 = 52(A – 1) = 3848
Hence a1 = 3848 – 3825 = 23