The number of point(s) satisfying the above mentioned characteristics and not in the first quadrant is/are

$$x_1 , x_2 , x_3,......., x_n$$ are in A.P. Let the first term be $$a$$ and common difference be $$d$$

Also, $$x_2 = a + d = 2$$

and $$x_{24} = a + 23d = 68$$

Solving above equations, we get : $$a = -1$$ and $$d = 3$$

=> x coordinates = {$$x_1 , x_2 , x_3, x_4 , x_5 ,.......$$} = {$${-1 , 2 , 5 , 8 , 11 , 14 ,......}$$}        -------------(i)

To find y coordinates, we will use $$y = px + q$$

$$\because A_2 (2, -2)$$ and $$A_{24} (68, 31)$$

Substituting in above equation, => $$-2 = 2p + q$$

and $$31 = 68p + q$$

Solving above equations, we get : $$p = \frac{1}{2}$$ and $$q = -3$$

=> $$y_1 = px_1 + q = \frac{1}{2}(-1) + (-3) = -3.5$$

Similarly, y coordinates = {$$y_1 , y_2 , y_3, y_4 , y_5 ,.......$$} = {$${-3.5 , -2 , \frac{-1}{2} , 1 ,......}$$}         -------(ii)

From (i) and (ii),

=> $$A_1 = (-1 , -3.5)$$

$$A_2 = (2 , -2)$$

$$A_3 = (5 , \frac{-1}{2})$$

$$A_4 = (8 , 1)$$

For the points to lie in the first quadrant, the coordinates of both $$(x_k , y_k)$$ must be positive.

Since, $$d$$ is positive and $$y$$ is a linear relation in $$x$$, the corresponding coordinates of $$A_k$$ i.e. $$A_5$$ onwards will be increasing.

Thus, only for $$A_1 , A_2$$ and $$A_3$$ we do not have both coordinates positive.

$$\therefore$$ Only 3 points do not lie in the first quadrant.