Let A = 1 , a 1 , a 2 … … a 18 , 77 be a set of integers with 1 < a 1 < a 2 < … . . < a 18 < 77 . Let the set A + A = x + y : x , y ∈ A contain exactly 39 elements. Then, the value of a 1 + a 2 + … . . + a 18 is equal to ______.

Question

Let A = 1 , a 1 , a 2 &#8230; &#8230; a 18 , 77 be a set of integers with 1 &#60; a 1 &#60; a 2 &#60; &#8230; . . &#60; a 18 &#60; 77 . Let the set A + A = x + y : x , y &#8712; A contain exactly 39 elements. Then, the value of a 1 + a 2 + &#8230; . . + a 18 is equal to ______.

Aicharya · Accepted Answer

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

Let A = 1 , a 1 , a 2 … … a 18 , 77 be a set of integers with 1 < a 1 < a 2 < &#8230…

Solution

Let A = 1 , a 1 , a 2 &#8230; &#8230; a 18 , 77 be a set of integers with 1 &#60; a 1 &#60; a 2 &#60; &#8230…

Solution

Let A = 1 , a 1 , a 2 … … a 18 , 77 be a set of integers with 1 < a 1 < a 2 < &#8230…