Question:

Let A = {1, a1, a2…a18, 77} be a set of integers with 1 <a1<a2<….<a18< 77. Let the set A + A = {x + y :x, y∈A} contain exactly 39 elements. Then, the value of a1 + a2 +…+a18 is equal to _____.

Updated On: Jun 29, 2025
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Correct Answer: 702

Solution and Explanation

If we write the elements of A + A, we can certainly find 39 distinct elements as 1 + 1, 1 + a1, 1 + a2,…..1 + a18, 1 + 77, a1 + 77, a2 + 77,……a18 + 77, 77 + 77.
It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.
Let the common difference be d.
\(77 = 1 + 19d\)
\(19d = 76\)
\(⇒d = 4\)
So,
\(\displaystyle\sum_{i=1}^{18} a_1\) \(= \frac {18}{2} [ 2a_1 + 17d ]\)
\(= 9 [ 10 + 68 ]\)
\(= 702\)

So, the answer is \(702\).

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Concepts Used:

Sets

In mathematics, a set is a well-defined collection of objects. Sets are named and demonstrated using capital letter. In the set theory, the elements that a set comprises can be any sort of thing: people, numbers, letters of the alphabet, shapes, variables, etc.

Read More: Set Theory

Elements of a Set:

The items existing in a set are commonly known to be either elements or members of a set. The elements of a set are bounded in curly brackets separated by commas.

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Cardinal Number of a Set:

The cardinal number, cardinality, or order of a set indicates the total number of elements in the set.

Read More: Types of Sets