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
1. 20
2. 23
3. 45
4. 48

Answer 1

Correct Answer - Option 2 : 23

Calculation:

As 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 subsumes 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 P.
Since a₅₂ = a₂₇ + 25 and a₅₂ = 100
⇒ P = 100 – 25 = 75
⇒ a₂ + a₃ + … + a₅₂ = 75 × 51 = 3825
Given a₁ + a₂ +… + a₅₂ = 52(P – 1) = 3848
Hence a₁ = 3848 – 3825 = 23

The average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term