Given \(a_3 = 4\) and \(a_9 = -8\). The formula for the nth term of an arithmetic progression is \(a_n = a + (n − 1) d\). Therefore, for the 3rd term, \(a_3 = a + (3 − 1) d\), which simplifies to \(4 = a + 2d\) (Equation i). For the 9th term, \(a_9 = a + (9 − 1) d\), which simplifies to \(-8 = a + 8d\) (Equation ii). Subtracting Equation (i) from Equation (ii) yields \(-12 = 6d\), so \(d = -2\). Substituting \(d = -2\) into Equation (i) gives \(4 = a + 2(−2)\), which simplifies to \(4 = a − 4\), thus \(a = 8\). Let the nth term of this A.P. be zero. Using the formula \(a_n = a + (n − 1) d\), we have \(0 = 8 + (n − 1) (−2)\). Simplifying this equation gives \(0 = 8 − 2n + 2\), leading to \(2n = 10\) and \(n = 5\).
Therefore, the 5th term of this A.P. is 0.