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.
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