Let the common ratio be r so that the first term is 2r. Sum, s = \(\frac{a}{1-r}\) ie. 8 = \(\frac{2r}{1-r}\) 8(1-r) = 2r, 8 - 8r = 2r 8 = 2r + 8r 8 = 10r r = \(\frac{4}{5}\). where common ratio (r) = \(\frac{second term(n_2)}{first term(a)}\), r = \(\frac{n_2}{2r}\) r = \(\frac{4}{5}\) and a = 2r or \(\frac{8}{5}\) \(\frac{4}{5}\) * \(\frac{8}{5}\) = n\(_2\) \(\frac{32}{25}\) = n\(_2\) The sum of the first two terms = a + n\(_2\) = \(\frac{8}{5}\) + \(\frac{32}{25}\) = \(\frac{40 + 32}{25}\) = \(\frac{72}{25}\)