OA Explanation:

Sum of infinite terms in a GP is given be, \(S_∞ = \frac{a}{(1 - r)}\), where a is the first term and r is the common ratio (r < 1)

\(12 = \frac{8}{(1 - r)}\)
\((1 - r) = \frac{8}{12} = \frac{2}{3}\)
i.e. \(r = \frac{1}{3}\)

nth term in GP is given by, \(T_n = ar^{n - 1}\)
\(T_4 = 8(\frac{1}{3})^{4 - 1}\)
\(T_4 = 8(\frac{1}{3})^3 = (\frac{8}{27})\)

Hence, option A