A sequence has a_1 = 9 and a_n = 3a_(n-1) - 20 for n >= 2. What is its fifth term?

1. Start from a_1 = 9, which the formula supplies. 2. a_2 = 3(9) - 20 = 27 - 20 = 7. 3. a_3 = 3(7) - 20 = 21 - 20 = 1. 4. a_4 = 3(1) - 20 = 3 - 20 = -17. 5. a_5 = 3(-17) - 20 = -51 - 20 = -71. 6. Stopping at a_4 gives -17 and forgetting to subtract 20 at the last step gives -51, so both are partial answers rather than the fifth term. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.1 Sequences and Their Notations_