If $9, x, y, z, a$ are in $A.P.$ such that $x + y + z = 15$,and $9, x, y, z, a$ are in $H.P.$ such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,then the value of $a$ is:

Let $A_1$ and $A_2$ be two arithmetic means and $G_1, G_2, G_3$ be three geometric means between two distinct positive numbers $a$ and $b$. Then $G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2$ is equal to

Let $S$ be the sum of the first $9$ terms of the series: $(x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + (x^4+(k+6)a) + \ldots$ where $a \neq 0$ and $x \neq 1$. If $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,then $k$ is equal to:

Let $a, b$ and $c$ be in $G.P.$ with common ratio $r,$ where $a \ne 0$ and $0 < r \le \frac{1}{2}.$ If $3a, 7b$ and $15c$ are the first three terms of an $A.P.,$ then the $4^{th}$ term of this $A.P.$ is

Let $x_{1}, x_{2}, x_{3}, \ldots, x_{20}$ be in a geometric progression with $x_{1} = 3$ and common ratio $r = \frac{1}{2}$. $A$ new data set is constructed by replacing each $x_{i}$ with $(x_{i} - i)^{2}$. If $\bar{x}$ is the mean of the new data,then the greatest integer less than or equal to $\bar{x}$ is $.....$

If the first three terms of the sequence $\frac{1}{16}, a, b, \frac{1}{6}$ are in a geometric progression and the last three terms are in a harmonic progression,then the values of $a$ and $b$ are: