If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

Let $a_{1}, a_{2}, \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}} = \frac{4}{9}$. If the sum of this $A.P.$ is $189$,then $a_{6} a_{16}$ is equal to:

$a_1, a_2, a_3, \dots, a_{100}$ are in an arithmetic progression,where $a_1 = 3$ and $S_p = \sum_{i=1}^p a_i, 1 \le p \le 100$. For any integer $n$,let $m = 5n$. If $S_m/S_n$ is independent of $n$,then $a_2 = \dots$

Six numbers are in an $AP$ such that their sum is $3$. The first term is $4$ times the third term. Then,the fifth term is:

Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms. Let $A_{k}=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2k-1}^2-a_{2k}^2$. If $A_3=-153$,$A_5=-435$ and $a_1^2+a_2^2+a_3^2=66$,then $a_{17}-A_7$ is equal to:

If the roots of the equation $6x^3-11x^2+6x-1=0$ are in harmonic progression,then the roots of $x^3-6x^2+11x-6=0$ will be in