Easy progression

Let $a$ be the first term, $n$ be the number of terms, $d$ be the common difference between the terms and $a_n$ be the last term.
So, $n = 3d\\ d = \dfrac{n}{3} \longrightarrow \boxed{1}$

$a_n = a + (n - 1)×d$
Substituting $\boxed{1}$ in the above equation, we get:-
$46 = 2 + (n - 1)×\dfrac{n}{3}\\ 44 = \dfrac{n^2 - n}{3}\\ 132 = n^2 - n\\ n^2 - n - 132 = 0\\ n^2 - \left(12n - 11n\right) - 132 = 0\\ n\left(n - 12\right) + 11\left(n - 12\right) = 0\\ \left(n - 12\right)\left(n + 11\right) = 0$
So, $\left(n - 12\right) = 0$ or $\left(n + 11 = 0\right)$ .
So, $n = 12$ or $n = -11$ .

Since number of yerms cannot be negative, $n \neq -11$ .
So, $n = \color{#69047E}{\boxed{12}}$ .

Using A.P.

We know $a_n=a+(n-1)d$ .

Given: $n=3d$ , $a=2$ and $l=46$ .

$\Rightarrow 46=2+(3d-1)d$

$3d^2-d-44=0$

$d=4,\dfrac{-11}{3}$ .

$\rightarrow d$ can't be $\dfrac{-11}{3}$ because number of terms cannot be negative.

$\therefore n=3d=\boxed{12}$