In a geometric progression, the sum of the first and the last terms is 66 and the product of the second and the last second terms is 128. Determine the first term of the series?

Let a be the first term and and r be the common ratio of the GP. From the given problem,
a + ar^{n - 1} = 66 ...(i)
Also, ar x ar^{n - 2} = 128 ⇒ a²r^{n - 1} = 128 ... (ii)
From Eq. (ii),
a.ar^{n - 1} = 128
ar^{n - 1} = 128/ a
On substitution this in Eq.(i), we get
a + 128/a = 66
a² - 66a + 128 = 0
a = [-b ± √b² - 4ac] /2a
= 66 ± √66² - 4 x 128 x 1)/2 = 64 or 2