A pack contains $n$ card numbered from $1$ to $n$ A pack contains $n$ card numbered from $1$ to $n$. Two consecutive numbered cards are removed from
the pack and sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on
the removing cards is $k$, Then $k$ is.
$\bf{My\; Try}::$ Let two consecutive cards be $k$ and $k+1,$ Then given sum of the number on the 
remaining cards is $1224$ .So $\left(1+2+3+.........+n\right)-\left(k+k+1\right) = 1224$
So $\displaystyle \frac{n(n+1)}{2}-(2k+1) = 1224\Rightarrow n(n+1)-(4k+2) = 2448$
Now I did not understand how can i calculate value of $(n,k)$
Help Required
Thanks 

You have $n^2+n-4k-2450=0$. Treat this as a quadratic in $n$ with constant term $-4k-2450$. Clearly we need a positive value, so
$$n=\frac{-1+\sqrt{1+16k+9800}}2=\frac{-1+\sqrt{16k+9801}}2\;,$$
$9801=99^2$; $100^2-99^2=199$, which is not a multiple of $16$, but $$101^2-99^2=2\cdot200=400=16\cdot25\;.$$
