Arithmetic means

Let $d$ be the common difference, $a$ be the first term and $l$ be the last term.
$d = \dfrac{l-a}{n+1} = \dfrac{6053-2}{n+1} = \dfrac{6051}{n+1}$
$A_1$ (first arithmetic mean) = $a + d = 2 + \dfrac{6051}{n+1} = \dfrac{2n + 6053}{n+1}$ .
$A_{n-5}$ (first arithmetic mean) = $a + (n-5)d = 2 + \dfrac{6051(n-5)}{n+1} = \dfrac{6053n - 30253}{n+1}$ .

$\dfrac{A_1}{A_{n-5}} = \dfrac{2n + 6053}{n+1} × \dfrac{n+1}{6053n - 30253} = \dfrac{1}{1207}\\ \dfrac{2n + 6053}{6053n - 30253} = \dfrac{1}{1207}\\ 1207\left(2n + 6053\right) = 6053n - 30253\\ 2414n + 7305971 = 6053n - 30253\\ 3639n = 7336224\\ n = \dfrac{7336224}{3639}\\ n = \color{#69047E}{\boxed{2016}}$ .