Question

Find the 8th term from the end of the GP 3, 6, 12, 24, ..., 12288.

Answer 1

Here r = 2 and 1 = 12288.

$\therefore$ 8th term from the end $=\frac{l}{r^{(8-1)}}=\frac{l}{r^7}=\frac{12288}{2^7}=\frac{12288}{128}=96$

Hence, the 8th term from the end = 96.