sequences and series - Problem about sum of arithemtic progression and geometric progression

Question: An arithmetic sequence has a common difference of $1$ and a geometric sequence has a common ratio of $3$. A new sequence is formed by adding corresponding terms of these two progressions. It is given that the second and fourth term of the sequence are $12$ and $86$ respectively. Find, in terms of $n$,

1. the $n^{th}$ term,


2. the sum of the first $n$ terms

of the new sequence.

Answers:

1. $n+1+3^n$.


2. $\frac 1 2 n(n+3) + \frac{3}{2} 3^n - \frac{3}{2}$.

Working so far

I am stuck getting different values for $a$.

Answer

The problem seems to be that you assume that the first term in both sequences is $a$.

Correct equations would be
$$u_2=a+1+3b=12$$
$$u_4=a+3+27b=86$$
where $a$ denotes the initial term of the arithmetic progression and $b$ is the initial term of the geometric progression.

I guess you can take it from here. (I got b=3 and a=2.)