1st term a = 3, 5th term = 9, sum of n = 81 nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9 3 + 4d = 9 4d = 9 - 3 d = \(\frac{6}{4}\) = \(\frac{3}{2}\) = 6 Sn = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\)) 81 = \(\frac{12n + 3n^2}{4}\) - 3n = \(\frac{3n^2 + 9n}{4}\) 3n2 + 9n = 324 3n2 + 9n - 324 = 0 By almighty formula positive no. n = 9 = 3