JEE Advanced · Mathematics · 5. Sequences & Series
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
- A 1
- B 3
- C 5
- D 7
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
Clearly, \(1+2+3+\ldots \ldots . .+ n -2 \leq 1224 \leq 3+4\) \(+\ldots \ldots . n\)
\(\Rightarrow \frac{( n -2)( n -1)}{2} \leq 1224 \leq \frac{( n -2)}{2}(3+ n ) \)
\( \Rightarrow n ^2-3 n -2446 \leq 0 \text { and } n ^2+ n~-\) \(2454 \geq 0 \)
\( \Rightarrow 49< n <51 \Rightarrow n =50 \)
\( \therefore \frac{ n ( n +1)}{2}-(2 k +1)=1224 \Rightarrow k =25 \Rightarrow k\) \(-~20=5\)
\(\Rightarrow \frac{( n -2)( n -1)}{2} \leq 1224 \leq \frac{( n -2)}{2}(3+ n ) \)
\( \Rightarrow n ^2-3 n -2446 \leq 0 \text { and } n ^2+ n~-\) \(2454 \geq 0 \)
\( \Rightarrow 49< n <51 \Rightarrow n =50 \)
\( \therefore \frac{ n ( n +1)}{2}-(2 k +1)=1224 \Rightarrow k =25 \Rightarrow k\) \(-~20=5\)
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