<s>[INST] An algebra question based on sequences and series The following question was asked in JEE Advanced 2014 Paper I:

A pack contains  $n$ cards numbered sequentially from $1$ to $n$. Two consecutive cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$. If the number on the smaller card is $k$, then what is the value of  $k-20$?

I could construct only one equation from this information:
$$\frac{n(n+1)}2-2k-1=1224$$
How should I proceed from here to solve this problem?
 [/INST] Let the two removed cards be $k$ and $k + 1$.
$1 + 2 + .... n = \frac {n(n+1)}2,$ and removing $k$ and $k+1$ and adding them up we get $\frac {n(n+1)}2 - k -(k+1) =\frac {n(n+1)}2 - 2k-1 = 1224,$
so $k = \frac {n(n+1)}4 - \frac {1225}2,$ which means $\frac {n(n+1)}2$ is odd and that
$\frac {n(n+1)}2 > 1225$.
So $n(n+1) > 2450 = 2500 -50 = 50(50-1) = 49*50$.
So $n > 49$.
We also have $n > k = \frac {n(n+1)}4 - \frac {1225}2$
$4n > n^2 + n - 2450 > 0$
$2450 > n^2 - 3n,$ so 
$2450+ \frac 94 > n^2 - 3n + \frac 92$
$\sqrt{2452.25}\approx 49.52 > n-\frac 32$
$51.02 > 51\ge n$
So $n = 50$ or $51$.  But we must have $\frac {n(n+1)}2$ be odd so $n= 50$ and
$k =\frac {50(51)}4 = \frac {1225}2 = \frac {25*51 - 1225}2=25$.
And $k -20 = 5.$
And indeed,  $(1+ ....+ 24) + (27+ ...+ 50) = (1+ .... + 50) - 25 - 26 = \frac {50*51}2 - 51 = 24*51 =  1224$.
.....
Argh.  Stupid errors trying to solve equations in one's head!
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