Multiple choice

Consider a sequence of 40 integers, denoted as a1, a2, ..., a40, where a1 is smaller than a2, a2 is smaller than a3, and so on in the increasing order. If the average of these 40 integers is three less than the average of the remaining integers a2, a3, ..., a40, and it is given that a40 equals 120, what is the maximum value that a1 can attain?

  1. 19

  2. -19

  3. -21

  4. 21
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let S be the sum of 40 integers. Average = S/40. Sum of remaining 39 = S - a1. Average of remaining = (S - a1)/39. Given S/40 = (S - a1)/39 - 3. Solving for a1 with a40=120 and the increasing order constraint leads to a1 = -19.

AI explanation

Let the average of the 40 integers be A, meaning their total sum is 40A. The average of the remaining 39 integers (a2 through a40) is A + 3, so their sum is 39(A + 3). Subtracting the first equation from the second gives the exact value of a1: 39(A + 3) - 40A = -A + 117. To find the maximum possible value of a1, we must minimize A. Since the sequence is strictly increasing and a40 = 120, the minimum possible average occurs if the integers are as small as possible while ending at 120, making a1 = -21. However, because a2 through a40 are 39 distinct integers whose average A must equal 117 - a1, setting a1 = -19 gives A = 136. This allows for a strictly increasing sequence ending at 120 (for example, centered around 136 with the upper values reaching 120), whereas a1 = -21 makes A = 138, requiring numbers far larger than 120, so the maximum possible value is -19.

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