If the first term of an infinite geometric series is twice the sum of all its subsequent terms,what is the common ratio?

There are two such pairs of non-zero real values of $a$ and $b$,i.e.,$(a_1, b_1)$ and $(a_2, b_2)$,for which $2a+b, a-b, a+3b$ are three consecutive terms of a $G.P.$. Then the value of $2(a_1b_2 + a_2b_1) + 9a_1a_2$ is-

Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a $G$.$P$. with common ratio $r = \frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \ldots + a_{10} = 62$, then $a_1$ is equal to:

The two geometric means between $1$ and $64$ are ........

If $a, b, c, d$ and $p$ are different real numbers such that $(a^{2}+b^{2}+c^{2}) p^{2}-2(ab+bc+cd) p+(b^{2}+c^{2}+d^{2}) \leq 0$,then show that $a, b, c$ and $d$ are in $G.P.$

If the first term of a $G.P.$ is $5$ and the common ratio is $-5$,then which term is $3125$ (in $^{th}$)?