Question

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 _____.

Answer 1

If we write the elements of A + A, we can certainly find 39 distinct elements as 1 + 1, 1 + a 1, 1 + a 2,…..1 + a 18, 1 + 77, a 1 + 77, a 2 + 77,…… 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.
$77 = 1 + 19d$
$19d = 76$
$⇒d = 4$
So,
$\displaystyle\sum_{i=1}^{18} a_1$ $= \frac {18}{2} [ 2a_1 + 17d ]$
$= 9 [ 10 + 68 ]$
$= 702$

So, the answer is $702$.