JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(3,7,11,15, \ldots, 403\) and \(2,5,8,11, \ldots, 404\) be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ...........
- A \(6696\)
- B \(6697\)
- C \(668\)
- D \(6699\)
Answer & Solution
Correct Answer
(D) \(6699\)
Step-by-step Solution
Detailed explanation
\(3,7,11,15, \ldots ., 403\) \(2,5,8,11, \ldots ., 404\) \(\operatorname{LCM}(4,3)=12\) \(11,23,35, \ldots . \text { let }(403)\) \(403=11+(n-1) \times 12\) \(\frac{392}{12}=n-1\) \(33 \cdot 66=n\) \(n=33\) \(\operatorname{Sum} \frac{33}{2}(22+32 \times 12)\) \(=6699\)
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