If   ⁿC₄, ⁿC₅ and ⁿC₆ are in A.P., then n can be :

If the sum of the first 15 terms of the series $${\left( {{3 \over 4}} \right)^3} + {\left( {1{1 \over 2}} \right)^3} + {\left( {2{1 \over 4}} \right)^3} + {3^3} + {\left( {3{3 \over 4}} \right)^3} + ....$$ is equal to 225 k, then k is equal to :

Let  S_k = $${{1 + 2 + 3 + .... + k} \over k}.$$ If   $$S_1^2 + S_2^2 + .....\, + S_{10}^2 = {5 \over {12}}$$A,  then A is equal to :

The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is :