$11^3 + 12^3 + .... + 20^3$

If the sum of the first $11$ terms of the series ${\left( {1\frac{4}{7}} \right)^2} + {\left( {1\frac{5}{7}} \right)^2} + {\left( {1\frac{6}{7}} \right)^2} + {2^2} + {\left( {2\frac{1}{7}} \right)^2} + ......$ is $\frac{{11}}{7}\lambda $,then $\lambda $ is equal to:

If $a^2 + b^2 + 16c^2 = 2(3ab + 6bc + 4ac)$,where $a, b, c$ are non-zero numbers,then $a, b, c$ are in:

Determine $k,$ so that $k+2, 4k-6$ and $3k-2$ are three consecutive terms of an $A.P.$

Consider a sequence whose sum of first $n$ terms is given by $S_n = 4n^2 + 6n$,where $n \in N$. Find the $15^{th}$ term $(T_{15})$ of this sequence.

If three positive numbers $a, b$ and $c$ are in $A.P.$ such that $abc = 8$,then the minimum possible value of $b$ is