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If I denotes the interest on a principal P (in dollars) at an interest rate of r per year for t years, then we have

I = Prt

The accumulated amount A, the sum of the principal and interest after t years is given by and is a linear function of t.

A= P + I = P + Prt = P(1 + rt)

A bank pays simple interest at the rate of 8% per year for certain deposits. a. If a customer deposits $1000 and makes no withdrawals for 3 years, what is the total amount on deposit at the end of three years?

P = 1000, r = 0.08, and t = 3 A=P(I+rt)=1000[1+(0.08)(3)]=1240 or $1240 b. What is the interest earned in that period? I=Prt=1000(0.08)(3)=240 or $240.

Compound interest Frequently, interest earned is periodically added to the principal and thereafter earns interest itself at the same rate. This is called compound interest. Suppose $1000 (the principal) is deposited in a bank for a term of 3 years, earning interest at the rate of 8% per year compounded annually. Using the simple interest formula we see that the accumulated amount

A1 = P (1+rt) =1000[1+0.08(1)] =1080

These observations suggest the following general rule:

If P dollars are invested over a term of t years earning interest at the rate of r per year compounded annually, then the accumulated amount is

This formula is derived under the assumption that interest was compounded annually.

In practice, however, interest is usually compounded more than once a year.

The interval of time between successive interest calculations is called the conversion period.

A=P(1+r)^t

i = r /m (annualinterest rate / Periods per year)

There are n = mt periods in t years, so the accumulated amount at the end of t years is given by

A=P[1+(r/m)]^n Where n = mt, and A= Accumulated amount at the end of t years P= Principal r= Nominal interest rate per year m= Number of conversion periods per year t= Term (number of years)

Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded

a. Annually b. Semiannually c. Quarterly . Daily

Annually, where, P = 1000, r = 0.08, and m = 1. Thus, i = r = 0.08 and n = 3, so A=1000[1 + (0.08/1)]^3 = 1259.71

b. Semiannually, Here, P = 1000, r = 0.08, and m = 2. Thus, i=0.08/2 and n = (3)(2) = 6, so A=1000[1+(0.08/2)]^6 =1265.32

c. Quarterly. P = 1000, r = 0.08, and m = 4.Thus, i=0.08/4 and n = (3)(4) = 12, so A=1000[1+(0.08/4)]^12 = 1268.24

d. Daily. P = 1000, r = 0.08, and m = 365. Thus,i=0.08/365 and n = (3)(365) = 1095, so A=1000[1+(0.08/365)]^1095 = 1271.22

Continuous Compound Interest Formula

A = Pe^rt where P= Principal r = Annual interest rate compounded continuously t= Time in years A= Accumulated amount at the end of t years

Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded (a) daily, and (b) continuously.

P = 1000, r = 0.08, m = 365, and t = 3, we find P = 1000, r = 0.08, and t = 3, we find

A = Pe^rt = 1000e^(0.08)(3) ≈ 1271.25

Note that the two solutions are very close to each other.

Effective Rate of Interest Formula r(eff) = [1+(r/m)]^m - 1 where reff = Effective rate of interest r= Nominal interest rate per year m= Number of conversion periods per year

Find the effective rate of interest corresponding to a nominal rate of 8% per year compounded a. Annually

Annually. Let r = 0.08 and m = 1. Then r(eff) = [1+(.08/1)]^1 - 1 =0.08 = 8%

If the effective rate of interest reff is known, then the accumulated amount after t years on an investment of P dollars can be more readily computed by using the formula

A=P(1+r (eff)) ^ t

Consider the compound interest formula:

A = P [1+(r/m)]^mt ---A -Future value and P -Present value

On occasion, investors may wish to determine how much money they should invest now, at a fixed rate of interest, so that they will realize a certain sum at some future date. This problem may be solved by expressing P in terms of A.

P=A(1+i)^ (-n) where i=r/m and n=mt

How much money should be deposited in a bank paying a yearly interest rate of 6% compounded monthly so that after 3 years the accumulated amount will be $20,000?

Here, A = 20,000, r = 0.06, m = 12, and t = 3.

P=A(1+(r/m)^(-mt) = 20000 [1+(.06/12)]^ (-12)(3) = 16713

How much money should be deposited in a bank paying a yearly interest rate of 6% compounded monthly so that after 3 years the accumulated amount will be $20,000?

Here, A = 20,000, r = 0.06, m = 12, and t = 3.

P=A(1+(r/m)^(-mt) = 20000 [1+(.06/12)]^ (-12)(3) = 16713

How long will it take $10,000 to grow to $15,000 if the investment earns an interest rate of 12% per year compounded quarterly?

Solution

A = 15,000, P = 10,000, r = 0.12, and m = 4, t=?

If the number of equations is greater than or equal to the number of variables in a linear system, then one of the following is true:

The system has no solution.

The system has exactly one solution.

The system has infinitely many solutions.

If there are fewer equations than variables in a linear system, then the system either has no solution or it has infinitely many solutions.…...

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