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

If the sum of the series $1 + \frac{2}{x} + \frac{4}{{{x^2}}} + \frac{8}{{{x^3}}} + ....\infty $ is a finite number, then

The solution of the equation $1 + a + {a^2} + {a^3} + ....... + {a^x}$ $ = (1 + a)(1 + {a^2})(1 + {a^4})$ is given by $x$ is equal to

Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to

The difference between the fourth term and the first term of a Geometrical Progresssion is $52.$ If the sum of its first three terms is $26,$ then the sum of the first six terms of the progression is

If the ${5^{th}}$ term of a $G.P.$ is $\frac{1}{3}$ and ${9^{th}}$ term is $\frac{{16}}{{243}}$, then the ${4^{th}}$ term will be