Mathematics - Sets and Relations Question with Solution | TestHub

If we write the elements of $A+A$, we can certainly find 39 distinct elements as $1+1,1+a_1,1+a_2,\dots .1$ $+a_{18},1+77,a_1+77,a_2+77,\dots \dots a_{18}+77,77+77$. 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 ' $d$ '. $77=1+19d\Rightarrow d=4$ So, ${\sum }_{i=1}^{18}{a}_1=\frac{18}{2}\left[2a_1+17d\right]=9[10+68]=702$