The first three terms of a geometric series are (3k+8)(3k+8)(3k+8), (k+8)(k+8)(k+8), and k k\,k respectively, where k k\,k is a positive constant.
Show that k2−4k−32=0k^2 - 4k - 32 = 0k2−4k−32=0.
Hence show that k=8k = 8k=8.
Find the common ratio.
Find the sum to infinity of the series.
