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Annuities and sinking funds

Section overview

In this section, you will learn to:

  1. Find the future value of an annuity.
  2. Find the amount of payments to a sinking fund.

In [link] and [link] , we did problems where an amount of money was deposited lump sum in an account and was left there for the entire time period. Now we will do problems where timely payments are made in an account. When a sequence of payments of some fixed amount are made in an account at equal intervals of time, we call that an annuity . And this is the subject of this section.

To develop a formula to find the value of an annuity, we will need to recall the formula for the sum of a geometric series.

A geometric series is of the form: a + ar + ar 2 + ar 3 + . . . + ar n size 12{a+ ital "ar"+ ital "ar" rSup { size 8{2} } + ital "ar" rSup { size 8{3} } + "." "." "." + ital "ar" rSup { size 8{n} } } {} .

The following are some examples of geometric series.

3 + 6 + 12 + 24 + 48 size 12{3+6+"12"+"24"+"48"} {}
2 + 6 + 18 + 54 + 162 size 12{2+6+"18"+"54"+"162"} {}
37 + 3 . 7 + . 37 + . 037 + . 0037 size 12{"37"+3 "." 7+ "." "37"+ "." "037"+ "." "0037"} {}

In a geometric series, each subsequent term is obtained by multiplying the preceding term by a number, called the common ratio. And a geometric series is completely determined by knowing its first term, the common ratio, and the number of terms.

In the example, a + ar + ar 2 + ar 3 + . . . + ar n 1 size 12{a+ ital "ar"+ ital "ar" rSup { size 8{2} } + ital "ar" rSup { size 8{3} } + "." "." "." + ital "ar" rSup { size 8{n - 1} } } {} the first term of the series is a size 12{a} {} , the common ratio is r size 12{r} {} , and the number of terms are n size 12{n} {} .

In your algebra class, you developed a formula for finding the sum of a geometric series. The formula states that the sum of a geometric series is

a r n 1 r 1 size 12{ { {a left [r rSup { size 8{n} } - 1 right ]} over {r - 1} } } {}

We will use this formula to find the value of an annuity.

Consider the following example.

If at the end of each month a deposit of $500 is made in an account that pays 8% compounded monthly, what will the final amount be after five years?

There are 60 deposits made in this account. The first payment stays in the account for 59 months, the second payment for 58 months, the third for 57 months, and so on.

The first payment of $500 will accumulate to an amount of $ 500 1 + . 08 / 12 59 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } } {} .

The second payment of $500 will accumulate to an amount of $ 500 1 + . 08 / 12 58 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"58"} } } {} .

The third payment will accumulate to $ 500 1 + . 08 / 12 57 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"57"} } } {} .

And so on.

The last payment is taken out the same time it is made, and will not earn any interest.

To find the total amount in five years, we need to add the accumulated value of these sixty payments.

In other words, we need to find the sum of the following series.

$ 500 1 + . 08 / 12 59 + $ 500 1 + . 08 / 12 58 + $ 500 1 + . 08 / 12 57 + . . . + $ 500 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"58"} } +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"57"} } + "." "." "." +$"500"} {}

Written backwards, we have

$ 500 + $ 500 1 + . 08 / 12 + $ 500 1 + . 08 / 12 2 + . . . + $ 500 1 + . 08 / 12 59 size 12{$"500"+$"500" left (1+ "." "08"/"12" right )+$"500" left (1+ "." "08"/"12" right ) rSup { size 8{2} } + "." "." "." +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } } {}

This is a geometric series with a = 500 size 12{a="500"} {} , r = 1 + . 08 / 12 size 12{r= left (1+ "." "08"/"12" right )} {} , and n = 59 size 12{n = "59"} {} . Therefore the sum is

$ 500 1 + . 08 / 12 60 1 . 08 / 12 = $ 500 73 . 47686 = $ 36738 . 43 size 12{ matrix { { {$"500" left [ left (1+ "." "08"/"12" right ) rSup { size 8{"60"} } - 1 right ]} over { "." "08"/"12"} } {} ## =$"500" left ("73" "." "47686" right ) {} ##=$"36738" "." "43" } } {}
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When the payments are made at the end of each period rather than at the beginning, we call it an ordinary annuity .

Future value of an ordinary annuity

If a payment of m size 12{m} {} dollars is made in an account n size 12{n} {} times a year at an interest r size 12{r} {} , then the final amount A size 12{A} {} after t size 12{t} {} years is

A = m 1 + r / n nt 1 r / n size 12{A= { {m left [ left (1+r/n right ) rSup { size 8{ ital "nt"} } - 1 right ]} over {r/n} } } {}
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Tanya deposits $300 at the end of each quarter in her savings account. If the account earns 5.75%, how much money will she have in 4 years?

The future value of this annuity can be found using the above formula.

= $ 300 1 + . 0575 / 4 16 1 . 0575 / 4 = $ 300 17 . 8463 = $ 5353 . 89 size 12{ matrix { = { {$"300" left [ left (1+ "." "0575"/4 right ) rSup { size 8{"16"} } - 1 right ]} over { "." "0575"/4} } {} ## =$"300" left ("17" "." "8463" right ) {} ##=$"5353" "." "89" } } {}
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Robert needs $5000 in three years. How much should he deposit each month in an account that pays 8% in order to achieve his goal?

If Robert saves x size 12{x} {} dollars per month, after three years he will have

x 1 + . 08 / 12 36 1 . 08 / 12 size 12{ { {x left [ left (1+ "." "08"/"12" right ) rSup { size 8{"36"} } - 1 right ]} over { "." "08"/"12"} } } {}

But we'd like this amount to be $5,000. Therefore,

x 1 + . 08 / 12 36 1 . 08 / 12 = $ 5000 x 40 . 5356 = $ 5000 x = 5000 40 . 5356 x = $ 123 . 35 size 12{ matrix { { {x left [ left (1+ "." "08"/"12" right ) rSup { size 8{"36"} } - 1 right ]} over { "." "08"/"12"} } =$"5000" {} ## x left ("40" "." "5356" right )=$"5000" {} ##x= { {"5000"} over {"40" "." "5356"} } {} ## x=$"123" "." "35"} } {}
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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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