QJEEM22Ma1518: JEE Main 2022 Mathematics - Class 11 JEEM22Ma1518 Solution: Let A = {1, a 1 , a 2 ....... a 18 , 77} be a set of integers…

If we write the elements of $A+A$ , 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 ' $d$ '.

$77=1+19 \mathrm{~d} \Rightarrow d=4$

So, $\sum\limits_{i=1}^{18} a_{1}=\frac{18}{2}\left[2 a_{1}+17 d\right]=9[10+68]=702$