The sum of the first $20$ terms of a $GP$ is $1025$ times the sum of its first $10$ terms. Find the common ratio of the $GP$.

Let $a_1, a_2, \dots, a_{49}$ be in $A.P.$ such that $\sum_{k=0}^{12} a_{4k+1} = 416$ and $a_9 + a_{43} = 66$. If $a_1^2 + a_2^2 + \dots + a_{17}^2 = 140m$,then $m = \dots$

If the $9^{th}$ term of an $A.P.$ is zero,then the ratio of its $29^{th}$ term to its $19^{th}$ term is:

The sum of the series $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots$ up to $15$ terms is

The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is

Two sequences ${t_n}$ and ${s_n}$ are defined by $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right)$ and $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$. Then: