the first three terms of an arithmethic series have a sum of 24 and a product of 312. What is the fourth term of the series?
The sum of an arithmetic series is given by:
$\displaystyle S_n = n\left(\frac{t_1+t_n}{2}\right)$
Knowing that, you can get from your given information that:
$\displaystyle t_1 + t_2 + t_3 = 24$
$\displaystyle 24 = 3\left(\frac{t_1+t_3}{2}\right)$
Solving both equations simultaneously yields the answer. For the second part, simply find the common difference and add it to the third term.
a + (a + d) + (a + 2d) = 24 => 3a + 3d = 24 => a + d = 8 => a = 8 - d .... (1)
a(a + d)(a + 2d) = 312 .... (2)
Substitute (1) into (2): (8 - d)(8)(8 + d) = 312 => (8 - d)(8 + d) = 39 => 64 - d^2 = 39 => d^2 = 25 => d = 5 or -5.
Case 1: d = 5 => a = 3.
Case 2: d = -5 => a = 13.
It's left for you to find the fourth term of the series in each case.