Nelson Functions 11
Nelson Functions 11
1st Edition
Chris Kirkpatrick, Marian Small
ISBN: 9780176332037
Table of contents
Textbook solutions

All Solutions

Section 8-5: Annuities: Present Value

Exercise 1
Step 1
1 of 5
a.)
(i) We need to draw the diagram containing 5 payments of $$650$ each and its corresponding present value as shown below.

(ii) The present value is therefore

$PV=650(1.037)^{-1}+650(1.037)^{-2}+650(1.037)^{-3}+…+650(1.037)^{-5}$

(iii) Using the formula for the present value of annuity

$PV=Rcdot dfrac{1-(1+i)^m}{i}=650times dfrac{1-1.037^{-5}}{0.037}=$2918.24$

(iv) The interest is the difference between the present value and the total amount paid.

$$
I=650(5)-2918.24=$331.77
$$

Exercise scan

Step 2
2 of 5
b.)

(i) We need to draw the diagram containing 18 payments of $$1200$ each and its corresponding present value as shown below.

(ii) The present value is therefore

$PV=1200(1.047)^{-1}+1200(1.047)^{-2}+1200(1.047)^{-3}+…+1200(1.047)^{-18}$

(iii) Using the formula for the present value of annuity

$PV=Rcdot dfrac{1-(1+i)^m}{i}=1200times dfrac{1-1.047^{-18}}{0.047}=$14,362.17$

(iv) The interest is the difference between the present value and the total amount paid.

$$
I=18(1200)-14,362.17=$7237.83
$$

Exercise scan

Step 3
3 of 5
c.)

(i) We need to draw the diagram containing $4times 3.5=14$ payments of $$84.73$ each and its corresponding present value as shown below.

(ii) The present value is therefore

$PV=84.73(1.009)^{-1}+84.73(1.009)^{-2}+84.73(1.009)^{-3}+…+84.73(1.009)^{-14}$

(iii) Using the formula for the present value of annuity

$PV=Rcdot dfrac{1-(1+i)^m}{i}=84.73times dfrac{1-1.009^{-14}}{0.009}=$1109.85$

(iv) The interest is the difference between the present value and the total amount paid.

$$
I=14(84.73)-1109.85=$76.37
$$

Exercise scan

Step 4
4 of 5
d.)

(i) We need to draw the diagram containing $12times 10=120$ payments of $$183.17$ each and its corresponding present value as shown below.

(ii) The present value is therefore

$PV=183.17(1.0055)^{-1}+183.17(1.0055)^{-2}+183.17(1.0055)^{-3}+…+183.17(1.0055)^{-120}$

(iii) Using the formula for the present value of annuity

$PV=Rcdot dfrac{1-(1+i)^m}{i}=183.17times dfrac{1-1.0055^{-120}}{0.0055}=$16,059.45$

(iv) The interest is the difference between the present value and the total amount paid.

$$
I=120(183.17)-16059.45=$5920.95
$$

Exercise scan

Result
5 of 5
See explanation inside.
Exercise 2
Step 1
1 of 5
The present value of the $m^{th}$ payment denoted as $PV_m$ is

$PV_m=R(1+i)^{-m}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

The sum of the series $PV_1+PV_2+PV_3+…+PV_m$ is the present value of the annuity which is calculated as

$$
PV=Rcdot dfrac{1-(1+i)^{-m}}{i}
$$

Step 2
2 of 5
a.)

(i) We shall calculate the present value of each payment
$(R=8000,; i=0.09,; m=7)$

$PV_n=R(1+i)^{-m}$

$PV_1=8000(1.09)^{-1}=$7339.45$

$PV_2=8000(1.09)^{-2}=$6733.44$

$PV_3=8000(1.09)^{-3}=$6177.47$

$PV_4=8000(1.09)^{-4}=$5667.40$

$PV_5=8000(1.09)^{-5}=$5199.45$

$PV_6=8000(1.09)^{-6}=$4770.14$

$PV_7=8000(1.09)^{-7}=$4376.27$

(ii) $PV = 8000(1.09)^{-1}+8000(1.09)^{-2}+8000(1.09)^{-3}+…+8000(1.09)^{-7}$

(iii) $PV=Rcdot dfrac{1-(1+i)^-m}{i}=8000cdot dfrac{1-1.09^{-7}}{0.09}=$40;263.62$

Step 3
3 of 5
b.)

(i) We shall calculate the present value of each payment
$(R=300,; i=0.04,; m=7)$

$PV_n=R(1+i)^{-m}$

$PV_1=300(1.04)^{-1}=$288.46$

$PV_2=300(1.04)^{-2}=$277.37$

$PV_3=300(1.04)^{-3}=$266.70$

$PV_4=300(1.04)^{-4}=$256.44$

$PV_5=300(1.04)^{-5}=$246.58$

$PV_6=300(1.04)^{-6}=$237.09$

$PV_7=300(1.04)^{-7}=$227.98$

(ii) $PV = 300(1.04)^{-1}+300(1.04)^{-2}+300(1.04)^{-4}+…+300(1.04^{-7}$

(iii) $PV=Rcdot dfrac{1-(1+i)^-m}{i}=300cdot dfrac{1-1.04^{-7}}{0.04}=$1800.62$

Step 4
4 of 5
c.)

(i) We shall calculate the present value of each payment
$(R=750,; i=0.02,; m=8)$

$PV_n=R(1+i)^{-m}$

$PV_1=750(1.02)^{-1}=$735.29$

$PV_2=750(1.02)^{-2}=$720.88$

$PV_3=750(1.02)^{-3}=$706.74$

$PV_4=750(1.02)^{-4}=$692.88$

$PV_5=750(1.02)^{-5}=$679.30$

$PV_6=750(1.02)^{-6}=$665.98$

$PV_7=750(1.02)^{-7}=$652.92$

$PV_8=750(1.02)^{-8}=$640.12$

(ii) $PV = 750(1.02)^{-1}+750(1.02)^{-2}+750(1.02)^{-3}+…+750(1.04)^{-8}$

(iii) $PV=Rcdot dfrac{1-(1+i)^-m}{i}=750cdot dfrac{1-1.02^{-7}}{0.02}=$5494.11$

Result
5 of 5
a.) $$40263.62$

b.) $$1800.62$

c.) $$5494.11$

Exercise 3
Step 1
1 of 6
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 6
a.)

We shall calculate the present value of the annuity
$(R=5000,; i=0.072,; m=5)$

$$
PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=8000cdot dfrac{1-1.072^{-5}}{0.072}=$20;391.67
$$

Step 3
3 of 6
b.)

We shall calculate the present value of the annuity
$(R=250,; i=0.024,; m=24)$

$$
PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=8000cdot dfrac{1-1.024^{-24}}{0.024}=$4521.04
$$

Step 4
4 of 6
c.)

We shall calculate the present value of the annuity
$(R=2550,; i=0.001,; m=100)$

$$
PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=8000cdot dfrac{1-1.001^{-100}}{0.001}=$2425.49
$$

Step 5
5 of 6
d.)

We shall calculate the present value of the annuity
$(R=48.50,; i=0.0195,; m=30)$

$$
PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=48.50cdot dfrac{1-1.0195^{-24}}{0.0195}=$1093.73
$$

Result
6 of 6
a.) $$20;391.67$

b.) $$4521.04$

c.) $$2425.49$

d.) $$1093.73$

Exercise 4
Step 1
1 of 3
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 3
We are given with present value $PV=1300$, $i=0.015$, $m=24$, and we shall calculate the regular payments.

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

$$
R=dfrac{PVcdot i }{1-(1+i)^{-m}}=dfrac{1300(0.015)}{1-1.015^{-24}}=$64.90
$$

Result
3 of 3
$$
$64.90
$$
Exercise 5
Step 1
1 of 6
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 6
$r=10%implies i=dfrac{0.1}{4}=0.025$

$t=4$ years, quarterly compounding $implies m=4times 4=16$

a.) The diagram is sketched below.

Step 3
3 of 6
Exercise scan
Step 4
4 of 6
b.) The present value of each R is $dfrac{R}{(1+i)^m}$

$7500=R(1.025)^{-1}+R(1.025)^{-2}+R(1.025)^{-3}+…+R(1.025)^{-16}$

Step 5
5 of 6
c.) Using the formula for the present value of annuity

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

$$
R=dfrac{PVcdot i}{1-(1+i)^{-m}}=dfrac{7500(0.025)}{1-1.025^{-16}}=$574.49
$$

Result
6 of 6
a.) see diagram

b.) $7500=R(1.025)^{-1}+R(1.025)^{-2}+R(1.025)^{-3}+…+R(1.025)^{-16}$

c.) $R=$574.49$

Exercise 6
Step 1
1 of 4
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 4
a.) In this case,

$R=40$ , $i=0.015$, $m=10$

The present value of the loaned amount is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=40cdot dfrac{1-1.015^{-10}}{0.015}=$368.89$

Since Rocco made an initial down payment of $$50$, the total present value is

$$
368.89+50=boxed{bf{$418.89}}
$$

Step 3
3 of 4
b.) The interest paid is the difference between the present value of the annuity and the actual amount paid.

Since there are 10 payments of $$40$ each, the interest is

$$
I=10(40)-368.89=boxed{bf{$31.11}}
$$

Result
4 of 4
a.) $$418.89$

b.) $$31.11$

Exercise 7
Step 1
1 of 3
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 3
a.) In this case,

$PV=128,000$

$i=0.0065$

$m=300$

The regular payments can be calculated as

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}implies R=dfrac{PVcdot i}{1-(1+i)^{-m}}$

$$
R=dfrac{128,000(0.0065)}{1-1.0065^{-300}}=boxed{bf{$971.03}}
$$

Result
3 of 3
$$
R=$971.03
$$
Exercise 8
Step 1
1 of 4
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 4
a.) We need to find the monthly payment for the 7-year loan.

$m=12cdot 7=84$

$i=0.12/12=0.01$

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

$R_7=dfrac{PVcdot i}{1-(1+i)^{-m}}=dfrac{64,000(0.01)}{1-1.01^{-84}}=boxed{bf{$1029.70}}$

For the 10-year loan, $m=10times 12=120$

$$
R_{10}=dfrac{64,000(0.01)}{1-1.01^{-120}}=boxed{bf{$810.72}}
$$

Step 3
3 of 4
b.) We shall calculate the interest which is the difference between PV and $mcdot R$

7-year loan: $I_7=mcdot R_7-PV=1029.70(84)-64000=$22,494.80$

10-year loan: $I_{10}=mcdot R_{10}-PV=$810.72(120)-64000=$33,286.40$

Thus, the 7-year loan would save her $33286.40-22294.80=boxed{bf{$10,791.60}}$

Result
4 of 4
a.) $R_7=$1029.70$ , $R_{10}=$810.72$

b.) 7-year loan would save her $$10,791.60$

Exercise 9
Step 1
1 of 3
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 3
We shall evaluate which of the two options is better. We shall compare the $R$ for both cases.

Option 1: Borrow $29000$ from the bank to pay the car in cash.

$PV=29000$ , $i=0.0045$ , $m=60$

$R_1=dfrac{PVcdot i}{1-(1+i)^{-m}}=dfrac{29000(0.0045)}{1-1.0045^{-60}}=$552.60$

Option 2: Finance at the dealership

$PV=32000$, $i=0.0020$ , $m=60$

$R_2=dfrac{32,000(0.002)}{1-1.002^{-60}}=$566.51$

Therefore, Option 1 is better since it requires lesser amount per month for the same period of 60 months.

Result
3 of 3
Borrow $$29000$ from the bank to pay the car in cash
Exercise 10
Step 1
1 of 4
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 4
a.) We need to find $R$ for each payment plan with different $m$

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}implies R=dfrac{PVcdot i}{1-(1+i)^{-m}}$

5-year plan: $m=5times 12=60 implies R=dfrac{35,000cdot 0.007}{1-1.007^{-60}}=$716.39$

10-year plan: $m=10times 12=120implies R=dfrac{35,000cdot 0.007}{1-1.007^{-120}}=$432.08$

15-year plan: $m=15times12=180implies R=dfrac{35,000cdot 0.007}{1-1.007^{-180}}=$342.61$

Step 3
3 of 4
b.) We shall calculate the interest

$I=Rcdot m – PV$

5-year plan: $I=(716.39)(60)-35000=$7983.40$

10-year plan: $I=432.08(120)-35000=$16849.60$

15-year plan: $I=342.61(180)-35000=$26669.80$

Result
4 of 4
a.) $R_5=$716.39$ , $R_{10}=$432.08$ , $R_{15}=$342.61$

b.) $I_5=$7983.40$ , $I_{10}=$16849.60$ , $I_{15}=$26669.80$

Exercise 11
Step 1
1 of 4
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 4
a.) Here, $R=25$ , $i=0.0155$ , $m=12$

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=25cdot dfrac{1-1.0155^{-12}}{0.0155}=$271.84$

Since $$45$ down payment is required, the price of the stereo is

$$
271.84+45=boxed{bf{$316.74}}
$$

Step 3
3 of 4
b.) The interest is calculated as

$$
I=Rcdot m – PV=25(12)-371.84=boxed{bf{$28.16}}
$$

Result
4 of 4
a.) $$271.84$

b.) $$28.16$

Exercise 12
Step 1
1 of 5
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 5
We’re given with

$PV=1800$ , $R=75.84$ and $m=30$, $n=12$, and we shall find $r$

$1800=75.84cdot dfrac{1-(1+r/12)^{-30}}{r/12}$

We shall solve this graphically.

Step 3
3 of 5
Exercise scan
Step 4
4 of 5
Therefore, $r=19%$
Result
5 of 5
$$
r=19%
$$
Exercise 13
Step 1
1 of 3
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 3
With $P=50,000$, quarterly compounding (n=4), and
$r=11.2%implies i=dfrac{0.112}{4}=0.028$, the account value after 20 years is

$A=P(1+i)^m=50,000(1+0.028)^{4(20)}=$455,427.52$

Now, we shall calculate how much quarterly withdrawals can it support for 10 years after retirement.

$m=10(4)=40$ , $i=0.028$

$R=dfrac{PVcdot i}{1-(1+i)^{-m}}=dfrac{455,427.42(0.028)}{1-(1.028)^{-40}}=boxed{bf{$19,070.96}}$

Therefore, Leo shall receive $$19,070.96$ every 3 months after retirement.

Result
3 of 3
$$
$19,070.96
$$
Exercise 14
Step 1
1 of 3
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 3
Charmaine plans on retiring 25 years from now and wants to have $R=$2500$ per month from her retirement account for 180 months.

$R=2500$

$i=0.0075$

$m=180$

The present value of such retirement plan is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=2500cdot dfrac{1-1.0075^{-180}}{0.0075}=$246,483.52$

We shall then find how much should she save every month over the next 25 years to reach her goal.

This time, $m=25times12=300$

$R=dfrac{PVcdot i}{1-(1+i)^{-m}}=dfrac{246,483.52cdot (0.0075)}{1-1.0075^{300}}=boxed{bf{$219.75}}$

Therefore, Charmaine needs to save $$219.75$ per month for 25 years.

Result
3 of 3
$$
$219.75
$$
Exercise 15
Step 1
1 of 6
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 6
a.) Here, we want to find $R$ such that $PV=660,000$ at $r=6.604%$ and $t=25$ years,

For weekly payment, $n=52$

$i=dfrac{0.06604}{52}=0.00127$

$m=52times 25=1300$

$R=dfrac{PVcdot i}{1-(1+i)^{-m}}=dfrac{660,000(0.00127)}{1-1.00127^{-1300}}=boxed{bf{$1037.45}}$

Thus, option B suits people who expects to live more than 25 years.

Step 3
3 of 6
b.) If $R=$1000$, we shall find the number of weeks can it support

$660,000=1000cdot dfrac{1-1.00127^{-m}}{0.00127}$

$1.00127^{-n}=0.1618$

Step 4
4 of 6
Exercise scan
Step 5
5 of 6
Option A can support 1435 weeks or $dfrac{1435}{52}=27.6$ years. Thus, option A suits for those who expect to live more than 27.6 years.
Result
6 of 6
a.) Option B: $$1037.45$ per week for 25 years

b.) Option A: $$1000.00$ per week for 27.6 years

Exercise 16
Step 1
1 of 2
Answers can vary.

a.) For instance, If you won a lottery, you will likely to get a lump sum payment and this does not earn interest once given unless you deposit or invest it somewhere that earns interest. If the prize is given as multiple payments, then it earns interest within the duration until the last payment.

b.) For instance, if you are planning for your retirement, the future value represents the amount you’ll be receiving during retirement, while the present value is the worth of investment that you have to save or accumulate today that can potentially earn interest and increases in value in the future.

Result
2 of 2
Answers can vary. See example inside.
Exercise 17
Step 1
1 of 5
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 5
a.) We’re given with

$R=500$

$i=0.0083$

$m=10$

$$
PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=500cdot dfrac{1-(1.083)^{-10}}{0.083}=boxed{bf{$3310.11}}
$$

Step 3
3 of 5
b.) $r=5.6%=0.056$ , $n=4$ (quarterly), $R=$500$, $t=6$ years

$i=r/n=0.056/4=0.014$

$m=ntimes t=4cdot 6=24$

$$
PV=Rcdot dfrac{1-(1+i)^{-m}}{i}=500cdot dfrac{1-1.014^{-24}}{0.014}=boxed{bf{$10,132.49}}
$$

Step 4
4 of 5
c.) You can simply substitute $A$ in the present value formula with the future value.
Result
5 of 5
a.) $$3310.11$

b.) $$10,132.49$

Exercise 18
Step 1
1 of 5
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 5
a.) We are given with

$R=325$

$PV=17000$

$i=0.006$

We shall find $m$

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

$17000=325cdot dfrac{1-1.006^{-m}}{0.006}$

$1.006^{-m}=1-dfrac{17000(0.006)}{325}$

$1.006^{-m}=dfrac{223}{325}$

Step 3
3 of 5
Exercise scan
Step 4
4 of 5
Therefore, it shall be paid completely in around 63 months.
Result
5 of 5
$63$ months
Exercise 19
Step 1
1 of 3
The present value of an annuity is

$PV=Rcdot dfrac{1-(1+i)^{-m}}{i}$

where

$R$ = regular payment

$i$ = interest rate per compounding period

$m$ = number of compounding periods

If the annual interest rate $r$, payment mode, and the number of years $t$ is given,

$i=dfrac{r}{n}$

$m=ntimes t$

$n$ = number of payment per year

Step 2
2 of 3
Let $Q$ be the regular payment done $q$ times while $Y$ is the regular payment done $y$ times.

We shall find the relationship between $Q$ and $Y$ such that the future value of $Q$ payments is equivalent to the present value of $Y$ payments for the same value of $i$.

$Qcdot dfrac{(1+i)^q-1}{i}=Ycdot dfrac{1-(1+i)^{-y}}{i}$

$Q=Ycdot dfrac{1-(1+i)^{-y}}{i} cdot dfrac{i}{(1+i)^q-1}$

$$
Q=Ycdot dfrac{1-(1+i)^{-y}}{(1+i)^q-1}
$$

Result
3 of 3
$$
Q=Ycdot dfrac{1-(1+i)^{-y}}{(1+i)^q-1}
$$
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Chapter 1: Introduction to Functions
Page 2: Getting Started
Section 1-1: Relations and Functions
Section 1-2: Function Notation
Section 1-3: Exploring Properties of Parent Functions
Section 1-4: Determining the Domain and Range of a Function
Section 1-5: The Inverse Function and Its Properties
Section 1-6: Exploring Transformations of Parent Functions
Section 1-7: Investigating Horizontal Stretches, Compressions, and Reflections
Section 1-8: Using Transformations to Graph Functions of the Form y 5 af [k(x 2 d)] 1 c
Page 78: Chapter Self-Test
Chapter 2: Equivalent Algebraic Expressions
Page 82: Getting Started
Section 2-1: Adding and Subtracting Polynomials
Section 2-2: Multiplying Polynomials
Section 2-3: Factoring Polynomials
Section 2-4: Simplifying Rational Functions
Section 2-5: Exploring Graphs of Rational Functions
Section 2-6: Multiplying and Dividing Rational Expressions
Section 2-7: Adding and Subtracting Rational Expressions
Page 134: Chapter Self-Test
Chapter 3: Quadratic Functions
Page 138: Getting Started
Section 3-1: Properties of Quadratic Functions
Section 3-2: Determining Maximum and Minimum Values of a Quadratic Function
Section 3-3: The Inverse of a Quadratic Function
Section 3-4: Operations with Radicals
Section 3-5: Quadratic Function Models: Solving Quadratic Equations
Section 3-6: The Zeros of a Quadratic Function
Section 3-7: Families of Quadratic Functions
Section 3-8: Linear-Quadratic Systems
Page 204: Chapter Self-Test
Page 206: Cumulative Review
Page 167: Check Your Understanding
Page 170: Practice Questions
Page 198: Check Your Understanding
Page 202: Practice Questions
Chapter 4: Exponential Functions
Page 212: Getting Started
Section 4-1: Exploring Growth and Decay
Section 4-2: Working with Integer Exponents
Section 4-3: Working with Rational Exponents
Section 4-4: Simplifying Algebraic Expressions Involving Exponents
Section 4-5: Exploring the Properties of Exponential Functions
Section 4-6: Transformations of Exponential Functions
Section 4-7: Applications Involving Exponential Functions
Page 270: Chapter Self-Test
Chapter 5: Trigonometric Ratios
Page 274: Getting Started
Section 5-1: Trigonometric Ratios of Acute Angles
Section 5-2: Evaluating Trigonometric Ratios for Special Angles
Section 5-3: Exploring Trigonometric Ratios for Angles Greater than 90°
Section 5-4: Evaluating Trigonometric Ratios for Any Angle Between 0° and 360°
Section 5-5: Trigonometric Identities
Section 5-6: The Sine Law
Section 5-7: The Cosine Law
Section 5-8: Solving Three-Dimensional Problems by Using Trigonometry
Page 340: Chapter Self-Test
Chapter 6: Sinusoidal Functions
Page 344: Getting Started
Section 6-1: Periodic Functions and Their Properties
Section 6-2: Investigating the Properties of Sinusoidal Functions
Section 6-3: Interpreting Sinusoidal Functions
Section 6-4: Exploring Transformations of Sinusoidal Functions
Section 6-5: Using Transformations to Sketch the Graphs of Sinusoidal Functions
Section 6-6: Investigating Models of Sinusoidal Functions
Section 6-7: Solving Problems Using Sinusoidal Models
Page 406: Chapter Self-Test
Page 408: Cumulative Review
Chapter 7: Discrete Functions: Sequences and Series
Page 414: Getting Started
Section 7-1: Arithmetic Sequences
Section 7-2: Geometric Sequences
Section 7-3: Creating Rules to Define Sequences
Section 7-4: Exploring Recursive Sequences
Section 7-5: Arithmetic Series
Section 7-6: Geometric Series
Section 7-7: Pascal’s Triangle and Binomial Expansions
Page 470: Chapter Self-Test
Chapter 8: Discrete functions: Financial Applications
Page 474: Getting Started
Section 8-1: Simple Interest
Section 8-2: Compound Interest: Future Value
Section 8-3: Compound Interest: Present Value
Section 8-4: Annuities: Future Value
Section 8-5: Annuities: Present Value
Section 8-6: Using Technology to Investigate Financial Problems
Page 536: Chapter Self-Test
Page 538: Cumulative Review