If the first term of an $A.P.$ is $3$ and the sum of its first $25$ terms is equal to the sum of its next $15$ terms,then the common difference of this $A.P.$ is:

If $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in $A.P.$, then $x$ equals:

If $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$,then $\sum_{r=1}^n \left[ \left( \prod_{i=1}^r x_i \right) - 2\sum_{i=1}^r (2i - 1) \right]$ is equal to:

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained will constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ will be formed again. Find the numbers.

$A$ person has two parents (father and mother),four grandparents,eight great-grandparents,and so on. Find the total number of ancestors the person has up to the $10^{th}$ generation.

If $3 + \frac{1}{4} (3 + d) + \frac{1}{4^2} (3 + 2d) + \dots \infty = 8$,then the value of $d$ is: