Precalculus, Chapter 9, 9.3, Section 9.3, Problem 29

You may use the following formula, such that:
a_n = a_1*r^(n-1)
Replacing 10 for n, 4 for a_1 and (1/2) for r, yields:
a_10 = 4*(1/2)^9 => a_10 = 2^(2-9) => a_10 = 2^(-7) => a_10 =1/128
Hence, the indicated term a_10 is a_10 =1/128.