Let A = {1, a_{1}, a_{2}, ldots, a_{18}, 77} | vedclass

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(C) Let the set $A$ have $n = 20$ elements. The set $A + A$ has $39$ elements. For a set with $n$ elements,the maximum number of elements in $A + A$ i

Vedclass · Accepted Answer

$702$

Vedclass · Answer

(C) Let the set $A$ have $n = 20$ elements. The set $A + A$ has $39$ elements. For a set with $n$ elements,the maximum number of elements in $A + A$ is $\frac{n(n+1)}{2} = \frac{20 	imes 21}{2} = 210$. The minimum number of elements is $2n - 1 = 2(20) - 1 = 39$. Since the set $A + A$ contains exactly $39$ elements,the set $A$ must be an arithmetic progression. Given $A = \{1, a_{1}, a_{2}, \ldots, a_{18}, 77\}$,the first term $a = 1$ and the last term $l = 77$ with $n = 20$ terms. The common difference $d$ is given by $77 = 1 + (20 - 1)d$,so $76 = 19d$,which implies $d = 4$. The terms are $1, 5, 9, 13, \ldots, 77$. The sum of the $18$ terms $a_{1} + a_{2} + \ldots + a_{18}$ is the sum of the arithmetic progression excluding the first and last terms. Sum $= \frac{18}{2} 	imes (a_{1} + a_{18}) = 9 	imes (5 + 73) = 9 	imes 78 = 702$.

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

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