Formula to find n where nth term >= x

Question

How many months does it take for $d$ to become at least $x$ where for $m$ months $a$ is added to $d$ $(d = a + d)$ but at every $m$+1 month $b$ is added to $d$ $(d = b + d)$ then repeats again while $d < x$?
For example: $d$ = 2, $a$ = 5, $m$ = 4, $b$ = 3, $x$ = 51 \begin{array}{| l | l | l | l |} \hline Month(n) & d & Increment\ amount & Final\ amount \\ \hline 1 & 2 & a & 2 + 5 = 7 \\ \hline 2 & 7 & a & 7 + 5 = 12 \\ \hline 3 & 12 & a & 12 + 5 = 17 \\ \hline 4 & 17 & a & 17 + 5 = 22 \\ \hline 5(m+1) & 22 & b & 22 + 3 = 25 \\ \hline 6 & 25 & a & 25 + 5 = 30 \\ \hline 7 & 30 & a & 30 + 5 = 35 \\ \hline 8 & 35 & a & 35 + 5 = 40 \\ \hline 9 & 40 & a & 40 + 5 = 45 \\ \hline 10(m+1) & 45 & b & 45 + 3 = 48 \\ \hline 11 & 48 & a & 48 + 5 = 53 \\ \hline \end{array} From the above table, it can be seen that $d$ $\geq$ $x$ at month 11. So it therefore took 11 months for $d$ to be more than or equal to $x$.
I was able to deduce that the above is a special case of arithmetic progression where the $n^{th}$ term can be found using a tweaked version of the arithmetic progression formula: $$ \alpha_n = \alpha_1 + ( n - 1)a - (floor(n \div m) \times ( a- b)) $$ where:
$\alpha_n = value\ of\ n^{th}\ term $
$\alpha_1 =\ first\ term $
$a =\ normal\ increment\ amount $
$b =\ increment\ amount\ to\ add\ at\ m\ +\ 1 $

Example: $d$ = 2, $a$ = 5, $m$ = 4, $b$ = 3, $x$ = 51, $n = 11$, $\alpha_1 = d+a $
$\alpha_n = \alpha_1 + ( n - 1)a - (floor(n \div m) \times ( a- b))$
$\alpha_{11}$ = $(d+a)$ + $(11-1)5$ - $(floor(11 \div 4) \times ( 5 - 3))$
$\alpha_{11}$ = $(2 + 5)$ + $(11-1)5$ - $(floor(11 \div 4) \times ( 5 - 3))$
$\alpha_{11}$ = $(7)$ + $(10)5$ - $(2 \times ( 2))$
$\alpha_{11}$ = $(7)$ + $50$ - $4$
$\alpha_{11}$ = $57$ - $4$
$\alpha_{11}$ = $53$

I was thinking the following equation can be used to find $n$ where $n$ can be transposed.
$$ \alpha_n = \alpha_1 + ( n - 1)a - (floor(n \div m) \times ( a- b)) $$ $$ \alpha_n \geq x $$ $$ \alpha_1 + ( n - 1)a - (floor(n \div m) \times ( a- b)) \geq x$$

Answer 1

Each block of five months adds $4a+b$. You need to add $x-d$, so the number of full five month blocks needed is $k=\lfloor \frac {x-d}{4a+b}\rfloor$ Now you are at $d+k(4a+b)$ so you need $x-d-k(4a+b)$. You get increments of $a$, so you need $\lceil \frac {x-d-k(4a+b)}a\rceil$ more months. We get to ignore the fact that the last month only gives us $b$ because we covered that in the first step. If you had a month before the last that gave $b$ we would need to be more careful.