JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Sequences and Series

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 =

Answer: 5

Solution

The sum of all

n

cards is

\frac{n ( n + 1 )}{2}

. After removing cards

k

and

k + 1

, the remaining sum is

\frac{n ( n + 1 )}{2} - (2 k + 1) = 1224

. Thus

\frac{n ( n + 1 )}{2} = 1225 + 2 k

. Since

k \geq 1

, we have

\frac{n ( n + 1 )}{2} \geq 1227

, so

n (n + 1) \geq 2454

. Trying

n = 49

49 \times 50 = 2450 < 2454

. Trying

n = 50

50 \times 51 = 2550

, giving

1275 = 1225 + 2 k

, so

2 k = 50

, thus

k = 25

. Therefore,

k - 20 = 5

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Integer answer

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