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 +

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 ______.

Quantrex Academy · Accepted Answer

Correct answer: 0

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+19 d d=4 ) So, ( _ i=1 ¹⁸ a₁= 18 2 [2 a₁+17 d ]=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 < … . . < 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 ______.

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More from Sets and Relations

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 ______.

Correct answer

Step-by-step solution

More from Sets and Relations

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 ______.