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 _____________.

Solution

Maths sir · 4 steps

Twenty numbers can never give fewer than 39 pairwise sums, so hitting exactly 39 forces an AP.

1. Fewest possible sums. A has 20 elements x₁ < x₂ < … < x₂₀. The chain x₁+x₁ < x₁+x₂ < … < x₁+x₂₀ < x₂+x₂₀ < … < x₂₀+x₂₀ already gives 20 + 19 = 39 different sums, so A+A always has at least 2n-1 = 39 elements.
2. Equality means AP. This minimum of 2n-1 is reached only when the elements are in arithmetic progression. Here A+A has exactly 39 elements, so 1, a₁, …, a₁₈, 77 is an AP with 20 terms.

   77 = 1 + (20-1)d ⇒ d = 76/19 = 4

3. Add the terms. The AP is 1, 5, 9, …, 77. Its 20 terms add to 20/2(1 + 77) = 780. Remove the two end terms: 780 - (1 + 77) = 702.
4. So. a₁ + a₂ + … + a₁₈ = 702.

The common slip: 780. Adding all 20 terms of the AP gives 780. The question wants only a₁ to a₁₈, so 1 and 77 must be removed.

The idea to keep. A set of n numbers has at least 2n-1 distinct pairwise sums x+y, and exactly 2n-1 only when the numbers form an AP. Count the elements first, then compare.

Answer from the NTA final answer key. Solution by PrepWiser.