Let A = 1, a 1, a 2 ldots ldots a 18, 77 be a set of integers with 1

Question

Let A = 1, a 1, a 2 ldots ldots a 18, 77 be a set of integers with 1 < a1 < a2 < ldots . < a18 < 77. Let the set A + A = x + y: x , y ∈ A contain exactly 39 elements. Then, the value of a1+a2+ ldots . .+a18 is equal to . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

a1, a2, a3, ldots, a18, 77 are in AP i.e. 1,5,9,13, ldots, 77 Hence a1+a2+a3+ ldots+a18=5+9+13+ ldots 18 terms =702

Q. Let $A = {1, a_{1}, a_{2} \dots\dots a_{18}, 77}$ be a set of integers with $1 < a_{1} < a_{2} < \dots . < a_{18} < 77$ . Let the set $A + A = {x + y : x, y \in A}$ contain exactly $39$ elements. Then, the value of $a_{1} + a_{2} + \dots .. + a_{18}$ is equal to ______.

$a_{1}, a_{2}, a_{3}, \dots, a_{18}, 77$
are in AP i.e. $1, 5, 9, 13, \dots, 77$
Hence $a_{1} + a_{2} + a_{3} + \dots + a_{18} = 5 + 9 + 13 + \dots 18$ terms
$= 702$

Q. 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∈A} contain exactly 39 elements. Then, the value of a1​+a2​+…..+a18​ is equal to ______.

a1​,a2​,a3​,…,a18​,77 are in AP i.e. 1,5,9,13,…,77 Hence a1​+a2​+a3​+…+a18​=5+9+13+…18 terms =702

Q. Let $A = {1, a_{1}, a_{2} \dots\dots a_{18}, 77}$ be a set of integers with $1 < a_{1} < a_{2} < \dots . < a_{18} < 77$ . Let the set $A + A = {x + y : x, y \in A}$ contain exactly $39$ elements. Then, the value of $a_{1} + a_{2} + \dots .. + a_{18}$ is equal to ______.

$a_{1}, a_{2}, a_{3}, \dots, a_{18}, 77$
are in AP i.e. $1, 5, 9, 13, \dots, 77$
Hence $a_{1} + a_{2} + a_{3} + \dots + a_{18} = 5 + 9 + 13 + \dots 18$ terms
$= 702$