1

### JEE Advanced 2013 Paper 1 Offline

Numerical
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 in 1224. If the smaller of the numbers on the removed cards is k, then k - 20 =

2

### IIT-JEE 2011 Paper 1 Offline

Numerical
Let $${{a_1}}$$, $${{a_2}}$$, $${{a_3}}$$........ $${{a_{100}}}$$ be an arithmetic progression with $${{a_1}}$$ = 3 and $${S_p} = \sum\limits_{i = 1}^p {{a_i},1 \le } \,p\, \le 100$$. For any integer n with $$1\,\, \le \,n\, \le 20$$, let m = 5n. If $${{{S_m}} \over {{S_n}}}$$ does not depend on n, then $${a_{2\,}}$$ is

3

### IIT-JEE 2010 Paper 1 Offline

Numerical
Let $${S_k}$$= 1, 2,....., 100, denote the sum of the infinite geometric series whose first term is $$\,{{k - 1} \over {k\,!}}$$ and the common ratio is $${1 \over k}$$. Then the value of $${{{{100}^2}} \over {100!}}\,\, + \,\,\sum\limits_{k = 1}^{100} {\left| {({k^2} - 3k + 1)\,\,{S_k}} \right|\,\,}$$ is

## Explanation

We have

$${S_k} = {{\left( {{{k - 1} \over {k!}}} \right)} \over {\left( {1 - {1 \over k}} \right)}} = {1 \over {(k - 1)!}}$$

Now, $$\sum\limits_{k = 2}^{100} {\left| {({k^2} - 3k + 1){1 \over {(k - 1)!}}} \right|}$$

$$= \sum\limits_{k = 2}^{100} {\left| {{{{{(k - 1)}^2} - k} \over {(k - 1)!}}} \right|}$$

$$= \sum {\left| {{{k - 1} \over {(k - 2)!}} - {k \over {(k - 1)!}}} \right|}$$

$$= \left| {{2 \over {1!}} - {3 \over {2!}}} \right| + \left| {{3 \over {2!}} - {4 \over {3!}}} \right| + ....$$

$$= {2 \over {1!}} - {1 \over {0!}} + {2 \over {1!}} - {3 \over {2!}} - {4 \over {3!}} + .... + {{99} \over {98!}} - {{100} \over {99!}}$$

$$= 3 - {{100} \over {99!}}$$

Thus, $${{{{100}^2}} \over {\left| \!{\underline {\, {100} \,}} \right. }} + \sum\limits_{k = 1}^{100} {\left| {({k^2} - 3k + 1){S_k}} \right| = 3}$$

4

### IIT-JEE 2010 Paper 2 Offline

Numerical
Let $${a_1},\,{a_{2\,}},\,{a_3}$$......,$${a_{11}}$$ be real numbers satisfying $${a_1} = 15,27 - 2{a_2} > 0\,\,and\,\,{a_k} = 2{a_{k - 1}} - {a_{k - 2}}\,\,for\,k = 3,4,........11$$. if $$\,\,\,{{a_1^2 + a_2^2 + a_{11}^2} \over {11}} = 90$$, then the value of $${{{a_1} + {a_2} + .... + {a_{11}}} \over {11}}$$ is equal to

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