Key Concepts and Formulas

• For a set $A$ of $n$ distinct integers, the set $A+A = \{x+y \mid x, y \in A\}$ has a minimum of $2n-1$ distinct elements.
• This minimum number of elements in $A+A$ is achieved if and only if the set $A$ forms an arithmetic progression (AP).
• The sum of an arithmetic progression is given by $S_k = \frac{k}{2}(a_1 + a_k)$, where $k$ is the number of terms, $a_1$ is the first term, and $a_k$ is the last term.

Step-by-Step Solution

Step 1: Determine the number of elements in set A and analyze the given condition. The set is given as $A = \{1, a_1, a_2, \ldots, a_{18}, 77\}$. The elements are ordered such that $1 < a_1 < a_2 < \ldots < a_{18} < 77$. The number of elements in set $A$ is $n = 1 + 18 + 1 = 20$. We are given that the set $A+A$ contains exactly 39 elements, i.e., $|A+A| = 39$.

Step 2: Deduce that set A must be an arithmetic progression. The minimum possible number of distinct elements in $A+A$ for a set $A$ with $n$ elements is $2n-1$. In this case, for $n=20$, the minimum number of elements in $A+A$ is $2(20)-1 = 40-1 = 39$. Since the given number of elements in $A+A$ is exactly 39, which is the minimum possible, the set $A$ must form an arithmetic progression.

Step 3: Determine the common difference of the arithmetic progression. Let the set $A$ be an arithmetic progression with the first term $x_1 = 1$ and the last term $x_{20} = 77$. The number of terms in the AP is $n=20$. The formula for the $n$-th term of an AP is $x_n = x_1 + (n-1)d$, where $d$ is the common difference. Substituting the known values: $x_{20} = x_1 + (20-1)d$ $77 = 1 + 19d$ Subtracting 1 from both sides: $76 = 19d$ Dividing by 19: $d = \frac{76}{19} = 4$ So, the common difference of the arithmetic progression is 4.

Step 4: Identify the elements $a_1, a_2, \ldots, a_{18}$ and calculate their sum. The elements of set $A$ are $x_1, x_2, \ldots, x_{20}$, where $x_1=1$ and $d=4$. The terms $a_1, a_2, \ldots, a_{18}$ correspond to $x_2, x_3, \ldots, x_{19}$ respectively. So, $a_i = x_{i+1} = x_1 + i \cdot d$ for $i = 1, 2, \ldots, 18$. The first term of the sequence $a_1, \ldots, a_{18}$ is $a_1 = x_2 = x_1 + 1 \cdot d = 1 + 1 \cdot 4 = 5$. The last term of this sequence is $a_{18} = x_{19} = x_1 + 18 \cdot d = 1 + 18 \cdot 4 = 1 + 72 = 73$. We need to find the sum $S = a_1 + a_2 + \ldots + a_{18}$. This is the sum of an arithmetic progression with 18 terms, where the first term is 5 and the last term is 73. Using the sum formula $S_k = \frac{k}{2}(\text{first term} + \text{last term})$: $S = \frac{18}{2}(a_1 + a_{18})$ $S = 9(5 + 73)$ $S = 9(78)$ $S = 702$

Common Mistakes & Tips

• Misinterpreting $|A+A|$ condition: The key to this problem is recognizing that $|A+A| = 2n-1$ implies $A$ is an AP. If this is missed, the problem becomes unsolvable.
• Off-by-one errors in indexing: Be careful when identifying $a_1, \ldots, a_{18}$ as specific terms within the full AP of $A$. They correspond to the 2nd through 19th terms of $A$.
• Incorrectly applying sum formula: Ensure the sum formula is applied to the correct sub-sequence ($a_1$ to $a_{18}$) with its specific first term, last term, and number of terms.

Summary

The problem leverages a fundamental property of set sums: for a set $A$ of $n$ integers, $|A+A| \ge 2n-1$, with equality holding if and only if $A$ is an arithmetic progression. Given $|A|=20$ and $|A+A|=39$, we deduce that $A$ is an AP. We then use the first and last elements of $A$ to find the common difference. Finally, we calculate the sum of the intermediate 18 terms of this AP.

The final answer is $\boxed{702}$.