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 :

If  a,   b,   c  are in A.P. and  a²,  b²,  c² are in G.P. such that
a < b < c and   a + b + c = $${3 \over 4},$$ then the value of a is :

If x₁, x₂, . . ., x_n and $${1 \over {{h_1}}}$$, $${1 \over {{h_2}}}$$, . . . , $${1 \over {{h_n}}}$$ are two A.P..s such that x₃ = h₂ = 8 and x₈ = h₇ = 20, then x₅.h₁₀ equals :

If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval :