The sums of $n$ terms of two arithmetic series are in the ratio $(2n + 3) : (6n + 5)$. Find the ratio of their $13^{th}$ terms.

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

The solution of $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$ is

The $9^{th}$ term of the series $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$ will be

If $y = x + x^2 + x^3 + \dots \infty$,then $x = $

If $x_1, x_2, \dots, x_n$ and $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$ are two $A.P.s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$,then $x_5 \cdot h_{10}$ equals