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 $\in$ A} contain exactly 39 elements. Then, the value of a1 + a2 + ...... + a18 is equal to _____________.
Answer (integer)
702
Solution
If we write the elements of $A+A$, we can certainly find 39 distinct elements as $1+1,1+a_{1}, 1+a_{2}, \ldots .1$ $+a_{18}, 1+77, a_{1}+77, a_{2}+77, \ldots \ldots a_{18}+77,77+77$.<br/><br/> 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.
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Let the common difference be '$d$'.
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$77=1+19 \mathrm{~d} \Rightarrow d=4$
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So, $\sum\limits_{i=1}^{18} a_{1}=\frac{18}{2}\left[2 a_{1}+17 d\right]=9[10+68]=702$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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