A pretty much random series

It's easy to tackle a problem by factoring out a common number to find a pattern. $2(1,2,5,x,41,122,365)$ Its easy to realise that the $n^{th}$ term is the sum of the previous term and $3^{n-2}$ . So, the original series would differ by $2 \times 3^{n-2}$

In this case, the 4th term would be $10 + 2 \times 3^{2}$ $\boxed{=28}$

find the difference from the previous number ,multiply it by 3 and add into the number will get a next number

4-2 =2, 2 *3 =6 , 6 +4 = 10

10-4 =6, 6 * 3 =18 , 18 + 10 =28

28-10 =18, 18 * 3 = 54 , 54 + 28 = 82

By keen observation, you can notice that the $x_{th}$ number in the pattern is: $3^{x-1}+1$ . Therefore the fourth number in the pattern is $3^{4-1}+1 = 27+1 = \boxed{28}$