Expection and Variance:
  If Y is a random variable with a geometric distribution,then
  $$E(Y) = \frac{1-p}{p}\;\;\;{\rm{and}}\;\;\;V(Y) = \frac{{1 - p}}{{{p^2}}}.$$
  You can use the following Mathematica command to obtain these results
    Expectation[x, x \[Distributed] GeometricDistribution[p]]
 or Mean[GeometricDistribution[p]]
    Variance[GeometricDistribution[p]]
A Important Property:
  Let Y denote a geometric random variable with probability of success p, $a$ is a nonnegative integer, then,
$$P(Y \ge a) = {(1 - p)^{a}}.$$
  For nonnegative integers $a$ and $b$,
$$P(Y  \ge  a + b|Y \ge a) = {(1 - p)^b} = P(Y \ge b).$$
  This property is called the memoryless property of the geometric distribution.