An algebra question based on sequences and series

Question

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?

Answer 1

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?

Answer 2

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!

Answer 3

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, $$k=25$$ $$\implies k-20=5$$