Let A = {1, a1, a2, ..., a18, 77} be a set of integers with 1

Q: Wait, why does $|A+A| = 2n-1$ force an Arithmetic Progression?

This is a classic result in additive combinatorics. If you have $n$ distinct integers, the smallest possible number of distinct sums $x+y$ you can form is $2n-1$. This minimum is achieved if and only if the set forms an arithmetic progression. If the set were not an A.P., there would be 'gaps' in the structure, leading to more than $2n-1$ distinct sums. It is a beautiful, rigid structure!

Q: Can't we just pick random numbers for $a_1, \dots, a_{18}$?

No, you cannot! The problem states that the sumset has *exactly* 39 elements. If you picked random numbers, the number of distinct sums would almost certainly be much larger than 39. The constraint $|A+A|=39$ is extremely restrictive and forces the set to be an arithmetic progression.

Q: Is there any other way to get 39 elements?

For a set of $n$ integers, the only way to have exactly $2n-1$ sums is if the set is an arithmetic progression. Any deviation from that structure increases the number of sums. So, no, there is no other configuration possible for this set.

Question

Let $A = \{1, a_1, a_2, \dots, a_{18}, 77\}$ be a set of integers with $1 < a_1 < a_2 < \dots < a_{18} < 77$. Let the set $A + A = \{x + y : x, y \in A\}$ contain exactly 39 elements. Then, the value of $a_1 + a_2 + \dots + a_{18}$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a set of twenty integers, $A = \{1, a_1, a_2, \dots, a_{18}, 77\}$. We know the smallest element is $1$ and the largest is $77$. There are eighteen mysterious numbers tucked in between, all strictly increasing.

At first glance, this looks like a problem of combinatorics, perhaps even a bit of chaos. But then, we are given a single, powerful clue: the sumset $A+A$, which contains all possible sums $x+y$ for $x, y \in A$, has exactly $39$ elements.

Why $39$? Let us look at the number of elements in $A$, which is $n=20$. If we calculate $2n-1$, we get $2(20)-1 = 39$. This is the smoking gun!

In the world of additive combinatorics, there is a fundamental theorem: for any set of $n$ integers, the size of the sumset $|A+A|$ is at least $2n-1$. When we see that our set hits this minimum bound exactly, it tells us something profound about the internal structure of $A$. It tells us that $A$ is not just a random collection of integers; it is a perfectly ordered, rigid structure—an **Arithmetic Progression**.

### Unlocking the Sequence

Once we realize that $A$ is an Arithmetic Progression, the problem transforms from a complex puzzle into a straightforward calculation. We have $n=20$ terms. The first term is $a=1$, and the twentieth term is $T_{20}=77$.

We need to find the common difference $d$. Using the standard formula for the $n$-th term of an A.P., $T_n = a + (n-1)d$, we can write:

$$77 = 1 + (20-1)d$$

This simplifies beautifully to $76 = 19d$. Dividing both sides by $19$, we find $d=4$. The gap between every consecutive term in our set is exactly $4$. Our set is $A = \{1, 5, 9, 13, \dots, 73, 77\}$.

### The Final Summation

Now, we are asked to find the sum of the eighteen middle terms: $a_1 + a_2 + \dots + a_{18}$. These are the terms from $5$ to $73$.

We have a sequence of $18$ terms where the first term is $5$ and the last is $73$. The sum of an arithmetic progression is given by the elegant formula $S = \frac{n'}{2}(	ext{first} + 	ext{last})$, where $n'$ is the number of terms. Substituting our values:

$$S = \frac{18}{2}(5 + 73)$$

$$S = 9 	imes 78$$

Calculating this, we get **$702$**. It is a satisfying conclusion to a journey that began with a mysterious set and ended with the precise, predictable beauty of an arithmetic progression.

Remember, in JEE problems, when you see a constraint that hits a theoretical minimum or maximum, it is almost always a hint about the underlying symmetry or structure of the system. Keep looking for that structure, and the math will always reveal itself.

Enter Numerical Value:

Visualized Solution

Understanding the Set $A$

The Sumset $A + A$

The Arithmetic Progression Theorem

Setting up the A.P. Equation

Solving for Common Difference $d$

Finding the Boundary Terms

Setting up the Sum Formula

Final Calculation

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

Unlocking the Sequence

The Final Summation

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