Question

Let $a_1, a_2, a_3, ..., a_n$ be a sequence of positive integers in arithmetic progression with common difference $d$ and $a_3 = 15$. Then, the maximum value of the sum $a_{a_1} + a_{a_2} + ... + a_{a_5}$ is ______.

Answer 1

495

Answer 2

The sequence is an arithmetic progression $a_1, a_2, ..., a_n$ with common difference $d$. The terms are given by $a_k = a_1 + (k-1)d$. We are given $a_3 = 15$, so $a_1 + 2d = 15$. Since $a_1$ is a positive integer, $a_1 \ge 1$, which implies $15 - 2d \ge 1$, so $2d \le 14$, and $d \le 7$.

The sum to maximize is $S = a_{a_1} + a_{a_2} + a_{a_3} + a_{a_4} + a_{a_5}$. The expression for the sum can be derived as $S = 5a_1(1+d) - 5d + 10d^2$. Substituting $a_1 = 15 - 2d$, we get $S = 5(15 - 2d)(1+d) - 5d + 10d^2 = 75 + 60d$.

We need to consider the constraint that $a_k$ must be positive integers for all $k \in \{1, 2, ..., n\}$. If $d > 0$, then $a_1 = 15 - 2d \ge 1$, so $d \in \{1, 2, 3, 4, 5, 6, 7\}$. In this case, all terms $a_k$ are positive. The sum $S = 75 + 60d$ is maximized when $d$ is maximized. The maximum value of $d$ is 7. For $d=7$, $a_1 = 15 - 2(7) = 1$. The sequence starts with $1, 8, 15, 22, 29, ...$. The indices $a_1, ..., a_5$ are $1, 8, 15, 22, 29$. The largest index is $a_5 = 29$, so $n \ge 29$. Since $a_1=1$ and $d=7$, all terms $a_k$ are positive. The maximum value of $S$ is $75 + 60(7) = 75 + 420 = 495$.

If $d = 0$, then $a_1 = 15$. The sequence is $15, 15, ...$. The sum is $S = 75 + 60(0) = 75$.

If $d < 0$, let $d = -|d|$. Then $a_1 = 15 + 2|d|$. For the terms $a_1, ..., a_5$ to be positive integers, we need $a_5 = 15 + (5-3)d = 15 - 2|d| > 0$, which means $2|d| < 15$, so $|d| \le 7$. However, for the sequence to consist of positive integers up to $a_n$, we must have $a_n > 0$. With $d < 0$, the terms decrease. We need $a_n = 15 + (n-3)d > 0$. Also, we need $n \ge \max(a_1, ..., a_5)$. The maximum index is $a_1 = 15 + 2|d|$. If $n \ge 15 + 2|d|$, then $a_n \ge 15 + (15 + 2|d| - 3)d = 15 + (12 + 2|d|)(-|d|) = 15 - 12|d| - 2|d|^2$. For $a_n > 0$, we need $2|d|^2 + 12|d| - 15 < 0$. The positive root of $2x^2 + 12x - 15 = 0$ is $x = \frac{-6 + \sqrt{66}}{2} \approx 1.05$. Since $|d|$ must be a positive integer, there is no integer $|d|$ such that $0 < |d| \le 7$ and $|d| < 1.05$. Thus, $d$ cannot be negative.

The possible integer values for $d$ are $\{0, 1, 2, 3, 4, 5, 6, 7\}$. The sum $S = 75 + 60d$ is maximized when $d=7$. The maximum value of the sum is $495$.