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?
 A: Hint:  Note that  $0 < k \le n-1$.
This means 
$${n (n+1) \over 2} > 1225$$
and $${n(n+1) \over 2} \le 1224 + 2n - 1 < 1224 + 2n.$$
Can you take it from here?
A: 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!
A: A/c to question
$${n(n+1)\over2} -(k)-(k+1)=1224$$
$$\implies k= n(n+1)-2450 /4$$
 Since $n(n+1)>2450$ & also k is integer then only $$n= 50$$ satisfies the above conditions.
Hence,
$$k=25$$
$$\implies k-20=5$$
