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← JEE Main 2022 · 28 Jun · Shift 1
MathematicsInteger+4 / -1Question 27 of 90

JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 27

Let A={1,a1,a2,…,a18,77}A=\{1,a_1,a_2,\dots,a_{18},77\} be a set of integers with 1<a1<a2<⋯<a18<771<a_1<a_2<\dots<a_{18}<77. Let the set A+A={x+y:x,y∈A}A+A=\{x+y:x,y\in A\} contain exactly 39 elements. Then, the value of a1+a2+⋯+a18a_1+a_2+\dots+a_{18} is equal to ______.

Answer and worked solution

Official answer: 702

Minimum size. For 20 integers, ∣A+A∣≥39|A+A|\ge39, with equality only for an arithmetic progression.

AP from 1 to 77 with 20 terms: d=4d=4. Sum of all terms =10⋅78=780=10\cdot78=780.

Middle terms. 780−78=702780-78=702.

Answer: 702