1. find the future value of the following:\n a. an ordinary annuity with p2,000 regular payment at 3%…

1. find the future value of the following:\n a. an ordinary annuity with p2,000 regular payment at 3% compounded annually for one year\n b. an ordinary annuity with p500 regular payment at 2% compounded quarterly for two years\n c. an annuity - due with p1,000 regular payment at 3% compounded monthly for one year\n d. an annuity - due of p2,000 regular payment at 2% compounded annually for 3 years\n2. in a savings account crediting 3% annual interest rate compounded monthly, p1,000 is deposited at the end of each month. what is the value of the account at the end of 18 months? how much interest is earned in all deposits?\n3. a 50 - year - old teacher wants to retire in 10 years. at his retirement he wants to treat his extended family to a vacation. he estimated that he needs p100,000 at the end of 10 years. in order to have this amount, he opened an account that gives 3% interest compounded annually. how much should he deposit in the account at the end of each year to accumulate p100,000 in 10 years?

1. find the future value of the following:\n a. an ordinary annuity with p2,000 regular payment at 3% compounded annually for one year\n b. an ordinary annuity with p500 regular payment at 2% compounded quarterly for two years\n c. an annuity - due with p1,000 regular payment at 3% compounded monthly for one year\n d. an annuity - due of p2,000 regular payment at 2% compounded annually for 3 years\n2. in a savings account crediting 3% annual interest rate compounded monthly, p1,000 is deposited at the end of each month. what is the value of the account at the end of 18 months? how much interest is earned in all deposits?\n3. a 50 - year - old teacher wants to retire in 10 years. at his retirement he wants to treat his extended family to a vacation. he estimated that he needs p100,000 at the end of 10 years. in order to have this amount, he opened an account that gives 3% interest compounded annually. how much should he deposit in the account at the end of each year to accumulate p100,000 in 10 years?

Answer

Explanation:

Step1: Recall the future - value formula for an ordinary annuity

The formula for the future - value of an ordinary annuity is $FVA = A\times\frac{(1 + r)^{n}-1}{r}$, where $A$ is the annuity payment, $r$ is the interest rate per period, and $n$ is the number of periods.

Step2: Solve part a

For an ordinary annuity with $A = 2000$, $r=0.03$ (annual interest rate), and $n = 1$ year. $FVA=2000\times\frac{(1 + 0.03)^{1}-1}{0.03}=2000\times\frac{1.03 - 1}{0.03}=2000\times\frac{0.03}{0.03}=2000$

Step3: Solve part b

The annual interest rate $i = 0.02$, compounded quarterly. So the interest rate per period $r=\frac{0.02}{4}=0.005$, and the number of periods $n = 2\times4 = 8$. The annuity payment $A = 500$. $FVA=500\times\frac{(1 + 0.005)^{8}-1}{0.005}$ First, calculate $(1 + 0.005)^{8}=1.005^{8}\approx1.040707$. Then $(1.005^{8}-1)\approx0.040707$. $FVA=500\times\frac{0.040707}{0.005}=500\times8.1414 = 4070.7$

Step4: Recall the future - value formula for an annuity - due

The formula for the future - value of an annuity - due is $FVAD=(1 + r)\times A\times\frac{(1 + r)^{n}-1}{r}$

Step5: Solve part c

The annual interest rate $i = 0.03$, compounded monthly. So the interest rate per period $r=\frac{0.03}{12}=0.0025$, and the number of periods $n = 1\times12 = 12$. The annuity payment $A = 1000$. $FVAD=(1 + 0.0025)\times1000\times\frac{(1 + 0.0025)^{12}-1}{0.0025}$ First, calculate $(1 + 0.0025)^{12}\approx1.030416$. Then $(1.030416 - 1)\approx0.030416$. $(1 + 0.0025)\times1000\times\frac{0.030416}{0.0025}=1.0025\times1000\times12.1664=12196.716$

Step6: Solve part d

The annual interest rate $i = 0.02$, the number of periods $n = 3$, and the annuity payment $A = 2000$. $FVAD=(1 + 0.02)\times2000\times\frac{(1 + 0.02)^{3}-1}{0.02}$ First, calculate $(1 + 0.02)^{3}=1.02^{3}=1.061208$. Then $(1.061208 - 1)=0.061208$. $(1 + 0.02)\times2000\times\frac{0.061208}{0.02}=1.02\times2000\times3.0604 = 6243.216$

Step7: Solve question 2

For the savings - account problem, $A = 1000$, the annual interest rate $i = 0.03$, compounded monthly. So $r=\frac{0.03}{12}=0.0025$, and $n = 18$ months. Using the ordinary - annuity formula $FVA = A\times\frac{(1 + r)^{n}-1}{r}$ $FVA=1000\times\frac{(1 + 0.0025)^{18}-1}{0.0025}$ $(1 + 0.0025)^{18}\approx1.045945$. Then $(1.045945 - 1)\approx0.045945$. $FVA=1000\times\frac{0.045945}{0.0025}=1000\times18.378 = 18378$ The total amount of deposits is $1000\times18=18000$. The interest earned is $18378 - 18000=378$

Step8: Solve question 3

Using the ordinary - annuity formula $FVA = A\times\frac{(1 + r)^{n}-1}{r}$, where $FVA = 100000$, $r = 0.03$, and $n = 10$. We need to solve for $A$: $100000=A\times\frac{(1 + 0.03)^{10}-1}{0.03}$ First, calculate $(1 + 0.03)^{10}\approx1.343916$. Then $(1.343916 - 1)\approx0.343916$. $A=\frac{100000\times0.03}{(1.03)^{10}-1}=\frac{3000}{0.343916}\approx8723.04$

Answer:

a. $2000$ b. $4070.7$ c. $12196.716$ d. $6243.216$ For question 2: The value of the account at the end of 18 months is $18378$, and the interest earned is $378$ For question 3: The teacher should deposit approximately $8723.04$ at the end of each year.