If in an infinite G.P., first term is equal to 10 times the sum of all successive terms,

(b) $\frac{1}{11}$

Let the first term of the G.P. be a.

Let its common ratio be r.

​According to the question, we have:

First term = 10        [Sum of all successive terms]

$a=10\left(\frac{a r}{1-r}\right)$

$\Rightarrow a-a r=10 a r$

$\Rightarrow 11 a r=a$

$\Rightarrow r=\frac{a}{11 a}=\frac{1}{11}$