If the sum of an infinite geometric series is $3$ and the sum of the squares of its terms is also $3$,what are the first term and the common ratio of the series?

Let $x_1, x_2, \ldots, x_6$ be the roots of the polynomial equation $x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$. Then,

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$ such that $B_n > A_n$ for all $n \geq p$ is:

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

Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.
$STATEMENT-1$ : The numbers $b_1, b_2, b_3, b_4$ are neither in $A.P.$ nor in $G.P.$
$STATEMENT-2$ : The numbers $b_1, b_2, b_3, b_4$ are in $H.P.$

Let the positive numbers $a_1, a_2, a_3, a_4$ and $a_5$ be in a $G$.$P$. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively,where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_3+a_4+a_5=14$,then $m+n$ is equal to $.........$.