Let $${1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,$$ (x_i $$ \ne $$ 0 for i = 1, 2, ..., n) be in A.P. such that x₁=4 and x₂₁ = 20. If n is the least positive integer for which $${x_n} > 50,$$ then $$\sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)} $$ is equal to :

Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.

$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series
1² + 2.2² + 3² + 2.4² + 5² + 2.6² ...........
If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to

Let    A_n = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3}$$ $$-$$. . . . . + ($$-$$1)^n-1 $${\left( {{3 \over 4}} \right)^n}$$    and    B_n = 1 $$-$$ A_n.
Then, the least dd natural numbr p, so that B_n > A_n , for all n$$ \ge $$ p, is :