Arithmetic progressions

Let $a$ be the first term, $a_2$ be the second term, $a_{n - 1}$ be the ${(n - 1)}^{\text{th}}$ term, $a_n$ be the $n^{\text{th}}$ term and $S_n$ be the sum of all terms of this arithmetic progression.
$a + a_{n - 1} = 991 \longrightarrow \boxed{1}\\ a_2 + a_n = 1005 \longrightarrow \boxed{2}$
Adding $\boxed{1}$ and $\boxed{2}$ , we get:-
$\color{#D61F06}{a} + \color{#3D99F6}{a_{n - 1}} + \color{#3D99F6}{a_2} + \color{#D61F06}{a_n} = 1996\\ 2\left(a + a_n\right) = 1996$
Because, sum of terms equidistant from the beginning and ending of an arithmetic progression are equal.
$a + a_n = 998 \longrightarrow \boxed{3}$ .

$S_n = \dfrac{n}{2} × \left(a + a_n\right)\\ 71357 = \dfrac{n}{2} × 998 \left(\text{From} \boxed{3}\right)\\ \dfrac{71357 × 2}{998} = n\\ n = \color{#69047E}{\boxed{143}}$