Question

Let $A =\left\{1, a _{1}, a _{2} \ldots \ldots a _{18}, 77\right\}$ be a set of integers with $1 < a_{1} < a_{2} < \ldots . < a_{18} < 77$. Let the set $A + A =\{ x + y : x , y \in A \} $ contain exactly $39$ elements. Then, the value of $a_{1}+a_{2}+\ldots . .+a_{18}$ is equal to ______.

Answer 1

$a_{1}, a_{2}, a_{3}, \ldots, a_{18}, 77$
are in AP i.e. $1,5,9,13, \ldots, 77$
Hence $a_{1}+a_{2}+a_{3}+\ldots+a_{18}=5+9+13+\ldots 18$ terms
$=702$