What is the probability that the first 1 comes in even trials of a six faced fair dice? A six faced fair dice is thrown until 1 comes, 
then what is the probability that 1 comes in even no. of trials?
I tried my best and my efforts include:
I find the
probability of getting a 1 on 2nd throw $p(2) = (5/6) (1/6)$
probability of getting a 1 on 4th throw $p(4) = (5/6)^3 (1/6)$
probability of getting a 1 on 6th throw $p(6) = (5/6)^5 (1/6)$
then I combine them like $p =  p(2)+p(4)+p(6)+ \ldots$ 
now, which formula should I apply to solve this problem?

Observe that if we define $\Pr(2k)$ to be the probability that the first 1 appears on the $2k$th throw, then 
$$\Pr(2k) = \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right)$$
Hence, the probability that the first one appears on an even-numbered throw is
$$\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right)$$
By extracting a factor of $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)$, we can obtain a convergent geometric series.
\begin{align*}
\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right) & = \frac{1}{6} \sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\\
& = \left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 2}\\
& = \frac{5}{36} \sum_{k = 1}^{\infty} \left[\left(\frac{5}{6}\right)^2\right]^{k -1}\\
& = \frac{5}{36} \sum_{k = 1}^{\infty} \left(\frac{25}{36}\right)^{k - 1}\\
& = \frac{5}{36} \cdot \frac{1}{1 - \frac{25}{36}} & \text{geometric series with $r = \frac{25}{36}$}\\
& = \frac{5}{36} \cdot \frac{1}{\frac{11}{36}}\\
& = \frac{5}{36} \cdot \frac{36}{11}\\
& = \frac{5}{11}
\end{align*}
