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The HBO co. is raising \$150,000 in 30 months by making monthly deposits which can be invested at 6%(12). Assuming they withdraw \$150,000 a month after the last deposit.

  1. If they deposit \$R per month for the first 10 months, \$2R for the next 10 months and \$3R for the last 10 months, what is R?

  2. If they deposit \$3R during the first 10 months, then \$2R during the second 10 months and \$R during the last 10 months, what is this time R?

  3. What is the reason of the difference in the answers of a. and b.?

I'm using the formula r = (s_ni)/[(1+i)^n-1] = (150,000(.06/12))/[(1+(.06/12)^10-1)]=14,665.59 but i don't think this is right.

manny singh
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1 Answers1

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HINT

(1) and (2) are just mechanical - compute the sums that accrue for each of the deposits, you get 3 geometric series.

for (3), the interest accrues differently. Thinmk about it on this simplified example. Would you prefer a bank account in which I deposit \$1 first and \$2 in a year, or the one with \$2 first and \$1 in a year?

gt6989b
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  • i did part 1 and 2 but i get same answer for both – manny singh Sep 26 '16 at 16:59
  • @mannyjim: you are probably not compounding the interest. If you use simple interest you will get the same answer, but in 2 the larger early deposits draw more interest. The first month is multiplied by $1.005^{29}≈1.33451.0129\approx 1.156$ while the last month is only multiplied by $1.005$. Clearly you want to multiply a large number by the larger amount. – Ross Millikan Sep 26 '16 at 17:16
  • Where u got 1.005^29 from? – manny singh Sep 26 '16 at 20:12
  • @mannyjim If you deposit $1 for 29 months at 5% interest, you get $1.05^{29}$ at the end – gt6989b Sep 26 '16 at 21:09