Suppose that $75000 is invested at 5 1/2% interest compounded quarterly

Question

Suppose that ?$75,000 is invested at 5 1/ 2?% ?interest, compounded quarterly. ?a) Find the function for the amount to which the investment grows after t years. ?b) Find the amount of money in the account at t=0?, 3?, 6?, and 10 years.

Answer 1

the interest rate (R) is always expressed as a yearly rate.

 

if compounded quarterly, each quarter amount of money increases by R/4

so it looks like this.

 

R   = Interest Rate per year 

M0 = Initial Amount at Time = 0

 

R = 5.5%  = 0.055  (expressed as a decimal)

each quarter (3 months) the rate applied (call it RQ) is R/4

 

RQ = 0.055/4 = 0.01375

 

after 1 quarter: 

M1  = M0 + M0*RQ   =  M0*(1 + RQ)  =  M0*(1.0375)

 

after two quarters: 

M2 = M1 + M1*RQ   = M1*(1.0375)   or   M0*(1.0375)(1.0375)

 

after 3 quarters:

M3 = M2 + M2*RQ   = M2(1.0375)     or   M0*(1.0375)(1.0375)(1.0375)     ok, see the pattern

 

after N quarters then, the amount M is then

#1)    M = M0*(1.0375)^N   or   M = M0*(1.0375)^t*4   (t=years)

 

 

OK, now we can solve for the amount M at 0, 3, 6, 10 years 

For each TIME in years, calculate N quarters, or just Years*4 

 

M0 = $75,000

 

for 0 years, N=0   

M(0) = M0*(1.0375)⁰     = M0*1  = $75,000  (note: any number to 0 exponent = 1)

 

at 3 years, N = 3*4 = 12 quarters

M(3 Years) = M0*(1.0375)¹²      = 75,000(1.0375)¹²  = $116,659.07

 

at 6 years, N = 6*4 = 24 quarters

M(6 Years) = M0*(1.0375)²⁴        = 75,000(1.0375)²⁴ = $181,457.86

 

at 10 years, N = 10*4 = 40 quarters

M(10 Years) = M0*(1.0375)⁴⁰    = 75,000(1.0375)⁴⁰ =  $327,028.41 

note: we expressed the equation 1 above with the quarterly interest rate (0.0375) and N = total number of months

 

The standard compounding formula is equivalent, which is:  

M = M0*(1 + R/X)^X*Y    Where R = yearly rate (as a decimal),

                                           X = number of times per year,

                                           Y = Years