What is the formula for compound interest (with n compounding periods per year)?

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Multiple Choice

What is the formula for compound interest (with n compounding periods per year)?

Explanation:
When interest is added more than once a year, the amount grows faster because each period applies interest to an already increased balance. The correct formula captures this by using the number of compounding periods per year. Here, P is the initial principal, r is the annual interest rate (as a decimal), n is how many times per year interest is credited, and t is the time in years. After nt compounding events, the balance becomes A = P(1 + r/n)^(nt). This structure makes sense: each period you multiply by 1 + r/n, and there are nt such periods. If you set n = 1, you get A = P(1 + r)^t, which is yearly compounding. In the limit as n grows without bound, this formula approaches A = P e^(rt), which is continuous compounding. Simple interest, A = P + rt, ignores compounding entirely, and A = P e^(rt) is the continuous-compounding case, not the discrete-period case. For example, with P = 1000, r = 0.05, n = 12, t = 3, A ≈ 1000(1 + 0.05/12)^(36) ≈ 1161.62, illustrating how increasing n increases the final amount.

When interest is added more than once a year, the amount grows faster because each period applies interest to an already increased balance. The correct formula captures this by using the number of compounding periods per year. Here, P is the initial principal, r is the annual interest rate (as a decimal), n is how many times per year interest is credited, and t is the time in years. After nt compounding events, the balance becomes A = P(1 + r/n)^(nt). This structure makes sense: each period you multiply by 1 + r/n, and there are nt such periods. If you set n = 1, you get A = P(1 + r)^t, which is yearly compounding. In the limit as n grows without bound, this formula approaches A = P e^(rt), which is continuous compounding. Simple interest, A = P + rt, ignores compounding entirely, and A = P e^(rt) is the continuous-compounding case, not the discrete-period case. For example, with P = 1000, r = 0.05, n = 12, t = 3, A ≈ 1000(1 + 0.05/12)^(36) ≈ 1161.62, illustrating how increasing n increases the final amount.

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