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=

Sequences and Series PYQ — JEE Advanced 2013 · Shift 1 · Q22 with Solution

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Correct answer: 5

Sum of all cards from $1$ to $n$

The total sum of numbers on cards from $1$ to $n$ is
$S_n = \frac{n(n+1)}{2}.$
Removed cards

Let the two consecutive removed cards be $k$ and $k+1$ .

Their sum is
$k + (k+1) = 2k+1.$
Condition from the question

After removing these two cards, the remaining sum is $1224$ :
$\frac{n(n+1)}{2} - (2k+1) = 1224.$
So,
$\frac{n(n+1)}{2} = 1225 + 2k.$
Look for a suitable triangular number near $1224$

Since $1225 = 35^2$ , check triangular numbers around this value:
$\frac{49\cdot 50}{2} = 1225.$
Thus if $n=49$ , total sum is $1225$ .

Then the removed sum would be
$1225 - 1224 = 1,$
which is impossible for two consecutive positive integers.

Next,
$\frac{50\cdot 51}{2} = 1275.$
Then the removed sum is
$1275 - 1224 = 51.$
Find $k$

Since
$2k+1 = 51,$
we get
$2k = 50 \implies k = 25.$
Therefore,
$k - 20 = 25 - 20 = 5.$
Final answer
$\boxed{5}$