JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(A =\left\{1, a _{1}, a _{2} \ldots \ldots a _{18}, 77\right\}\) be a set of integers with \(1< a _{1}< a _{2}<\ldots \ldots< a _{18}<77\). Let the set \(A + A =\{ x + y : x , y \in A \} \quad\) contain exactly \(39\) elements. Then, the value of \(a_{1}+a_{2}+\ldots \ldots+a_{18}\) is equal to...........
- A \(802\)
- B \(72\)
- C \(702\)
- D \(102\)
Answer & Solution
Correct Answer
(C) \(702\)
Step-by-step Solution
Detailed explanation
\(a _{1}, a _{2}, a _{3}, \ldots, a _{18}, 77\) are in \(AP\) i.e. \(1,5,9,13, \ldots, 77\). Hence \(a_{1}+a_{2}+a_{3}+\ldots+a_{18}=5+9+13+\ldots 18\) terms \(=702\)
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