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 =

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 $$k-20=$$

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

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