Sequences and Series 2 Question 21
22. 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 $k-20$ is equal to
(2013 Adv.)
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Solution:
- Let number of removed cards be $k$ and $(k+1)$.
$\therefore \frac{n(n+1)}{2}-k-(k+1)=1224$
$\Rightarrow \quad n^{2}+n-4 k=2450 \Rightarrow n^{2}+n-2450=4 k$
$\Rightarrow \quad(n+50)(n-49)=4 k$
$\therefore \quad n>49$
Let $\quad n=50$
$\therefore \quad 100=4 k$
$\Rightarrow \quad k=25$
Now $\quad k-20=5$
