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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2013 · Paper 1 · Mathematics · Sequences and Series

A pack contains n cards numbered from 1 to n. Two consecutive...

A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k−20=\text{k} - 2 0 =

Correct answer

5

Step-by-step solution

Clearly, 1 + 2 + 3+.......+n - 2 ≤\le 1224 ≤\le 3 + 4+.......n
⇒(n−2)(n−1)2≤1224≤(n−2)2(3+n)⇒ \dfrac{\left(\text{n} - 2\right) \left(\text{n} - 1\right)}{2} \le 1 2 2 4 \le \dfrac{\left(\text{n} - 2\right)}{2} \left(3 + \text{n}\right)
⇒n2−3n−2446≤0andn2+n−2454≥0⇒ {\text{n}}^{2} - 3 \text{n} - 2 4 4 6 \le 0 \text{and} {\text{n}}^{2} + \text{n} - 2 4 5 4 \ge 0
⇒49<n<51⇒n=50⇒ 4 9 < \text{n} < 5 1 ⇒ \text{n} = 5 0
∴n(n+1)2−(2k+1)=1224⇒k=25⇒k−20=5∴ \dfrac{\text{n} \left(\text{n} + 1\right)}{2} - \left(2 \text{k} + 1\right) = 1 2 2 4 ⇒ \text{k} = 2 5 ⇒ \text{k} - 2 0 = 5
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