Question 5.P.9: In Problem 5.8, suppose Mrs. Carter deposits $100 a month du......
In Problem 5.8, suppose Mrs. Carter deposits $100 a month during the first year, $110 a month during the second year, $120 a month during the third year, etc. How much will have accumulated at the end of 5 years if the interest rate is 6% per year, compounded continuously? (Compare Problem 4.12.)
Proceeding as in Problem 4.12, we have:
F = $100[F/A, 0.5%, 12] [F/P, 6%, 4] + $110[F/A, 0.5%, 12] [F/P, 6%, 3]
+ $120[F/A, 0.5%, 12] [F/P, 6%, 2] + $130[F/A, 0.5%, 12] [F/P, 6%, 1]
+ $140[F/A, 0.5%, 12]
The required numerical values can be obtained from Appendix C. Thus,
F = [$100(1.2712) + $110(1.1972) + $120(1.1275) + $130(1.0618) + $140](12.3364)
= ($672.146)(12.364) = $8291.86
A neater solution procedure (which might also have been applied in Problem 4.12) is to consider the deposits as constituting a five-year gradient series, with
A_o = $100[F/A, 0.5%, 12] and G = $10[F/A, 0.5%, 12]
Thus, as in Problem 5.6,
F = {$100[F/A, 0.5%, 12] + $10[F/A, 0.5%, 12] [A/G, 6%, 5]} [F/A′, 6%, 5]
= {$100 + $10[A/G, 6%, 5]} [F/A, 0.5%, 12] [F/A′, 6%, 5]
= [$100 + $10(1.8802)](12.3364)(5.6578) = $8291.87