T_{n} = 324, S_{n} = 484, r = 3, a =?, n =?

Now, S_{n} = \(\frac{rT _n−a}{r−1}\)

∴ 484 = \(\frac{3(324)−a}{3−1}\)

∴ 484 = \(\frac{972−a}{2}\)

∴ 484 × 2 = 972 – a

∴ a = 972 – 968

a = 4

Now, T_{n} = a.r^{n-1}

∴ 324 = 4(3)^{n-1}

∴ 324 / 4 = (3)^{n-1}

∴ (3)^{4} = (3)^{n-1}

∴ 4 = n – 1 ( Base are equal, Index are also equal)

∴ n = 5