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. if a₅₂= 100, then the largest possible value of a₁ is

a. 20
b. 23
c. 45
d. 48

Answer 1

Considering the conditions given and that the sequence of integers is strictly increasing, we determine that the largest possible value for a1, which allows the arithmetic mean to increase by 1 when it's removed, is 48. The correct answer is option d.

Step-by-step explanation:

Let's consider the arithmetic mean of the first sequence, a1, a2, ..., a52 which is one less than the arithmetic mean of the second sequence, a2, a3, ..., a52. This means the absence of a1 from the second sequence raises the mean by 1. Since we know that a52 = 100 and the arithmetic mean increases by removing the smallest number, we can find the largest possible value of a1 by making a2 through a51 as close to 100 as possible.

Since there are 52 numbers in the first sequence and 51 in the second, the sum of the second sequence is 51 more than the sum of the first sequence. This indicates that a1 = 100 - 51 = 49. But since our numbers must be positive integers and strictly increasing, a1 cannot be 49 as this would require a2 to also be 49, which violates the condition a1 < a2. Therefore, a1 must be one less, which is 48. So, the largest possible value of a1 is 48, making the correct option (d).