Points on Line in Arithmetic Progression

$X_2 = 2$ and $X_{24} = 68 \implies X_2 + 22d_x = 68 \implies 22d_x = 66 \implies d_x = 3$. $X_1 = -1$, $X_8 = -1 + 7(3) = 20$. $Y_2 = -2$ and $Y_{24} = 31 \implies -2 + 22d_y = 31 \implies 22d_y = 33 \implies d_y = 1.5$. $Y_1 = -3.5$, $Y_8 = -3.5 + 7(1.5) = 7$.

For a point $(X_k, Y_k)$ to lie in the first quadrant, both $X_k > 0$ and $Y_k > 0$. $A_1 = (-1, -3.5) \implies$ Quadrant III (not Q1). $A_2 = (2, -2) \implies$ Quadrant IV (not Q1). $A_3 = (5, -0.5) \implies$ Quadrant IV (not Q1). $A_4 = (8, 1) \implies$ Quadrant I. Since both $d_x = 3 > 0$ and $d_y = 1.5 > 0$, all subsequent points $A_5, A_6...$ have positive coordinates. Thus, exactly 3 points ($A_1, A_2, A_3$) do not lie in the first quadrant.