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JEE Main 202228 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
28 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2022 · 28 June · Morning Shift · Mathematics · Sets and Relations

Let A = \ 1, a _ 1 , a _ 2 a _ 18 , 77\ be a set of integers with 1...

Let A={1,a1,a2……a18,77}A = \left\{1, {a}_{1}, {a}_{2} \ldots \ldots {a}_{18}, 77\right\} be a set of integers with 1<a1<a2<…..<a18<771 < {a}_{1} < {a}_{2} < \ldots . . < {a}_{18} < 77. Let the set A+A={x+y:x,y∈A}A + A = \left\{x + y: x, y \in A\right\} contain exactly 3939 elements. Then, the value of a1+a2+…..+a18{a}_{1} + {a}_{2} + \ldots . . + {a}_{18} is equal to ______.

Correct answer

0

Step-by-step solution

If we write the elements of A+AA+A, we can certainly find 39 distinct elements as (1+1,1+a_1, 1+a_2, \ldots .1) (+a_{18}, 1+77, a_1+77, a_2+77, \ldots \ldots a_{18}+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 ' dd '. (77=1+19 d \Rightarrow d=4) So, (\sum_{i=1}^{18} a_1=\frac{18}{2}\left[2 a_1+17 d\right]=9[10+68]=702)
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