If we write the elements of A+A, we can certainly find 39 distinct elements as 1+1,1+a1,1+a2,….1 +a18,1+77,a1+77,a2+77,……a18+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⇒d=4 So, i=1∑18a1=218[2a1+17d]=9[10+68]=702