## College Algebra 7th Edition

The first five terms are: $a_1 = 0 \\a_2 = \ln{5} \\a_3 = 2\ln{5} \\a_4=3\ln{5} \\a_5=4\ln{5}$ The sequence is not geometric.
To find the first five terms, substitute 1, 2, 3, 4, and 5 to the given formula. Use the rules (1) $\ln{(a^n)} = n\cdot \ln{a}$ (2) $\ln{1} = 0$ $a_1 = \ln{(5^{1-1})}=\ln{(5^0)}=\ln{1}=0 \\a_2 = \ln{(5^{2-1})} = \ln{(5^1)} =\ln{5} \\a_3 = \ln{(5^{3-1})} = \ln{(5^2)} =2\ln{5} \\a_4=\ln{(5^{4-1})} = \ln{(5^3)} =3\ln{5} \\a_5=\ln{(5^{5-1})} = \ln{(5^4)} =4\ln{5}$ RECALL: A sequence is geometric if there is a common ratio among consecutive terms. Note that consecutive terms do not have a common ratio. Thus, the sequence is not geometric.