Question

A man deposits P650 every end of 6 months in an account paying 5 1/2 % interest compounded semiannually. What amount is in the account at the end of 5 years and 6 months?

Answer 1

Formula you're going to want to use;
\[A=P(1+\frac{ r }{ n })^{nt}\]

Answer 2

\[P=650\]
So install "P" into the equation;
\[A=650(1+\frac{ r }{ n })^{nt}\]
Then Install your interest rate;
\[A=650(1+\frac{ 5~1/2 percent }{ ? })^{nt}\]
Then Install the Time interest applied per month;
\[A=650(1+\frac{ 5~1/2 }{ 0.91 })^{(0.91)(t)}\]
Then install the time-period of elapsed time;
\[A=650(1+\frac{ 5~1/2 }{ 0.91 })^{(0.91)(5)}\]
Use the numbers according to get your final answer.

Answer 3

1 sec, where did .91 come from? could you explain I can't quite follow

Answer 4

dividing how much it gets per month, in a matter of 6 months, it goes up 5 1/2 percent. So, divide it into 6 to get your monthly average.

Answer 5

You don't know what n is then, its number of times compounded per year or month you choose,

Answer 6

Hi, the question requires you to find the future value of an ordinary annuity. It's an ordinary annuity because it's a series of payments made at the end of each period.
FV=pmt[((1+r)^n)-1]/r
Fv is the future value which we are looking for

Pmt is the periodic payment=650

r is the rate per period. Since interest is compounded semi annually, r=0.055/2=0.0275

n is the total number of deposits =11
(2 deposits per year, for 5 years and 6 months)

Substituting
FV=650*[((1+0.0275)^11)-1]/0.0275
FV=650*[(1.0275^11)-1]/0.0275
FV=650*(1.34772-1)/0.0275
FV=8218.87

The account will have P8218.87 at the end of 5 years and 6 months

Answer 7

\[amount=650(1.0275)^{11}+650(1.0275)^{10}+...+650(1.0275)^1\]
\[=650(1.0275)\frac{ 1.0275^{11}-1 }{ 1.0275-1}\approx 650(1.0275)\frac{ 1.34772143598-1 }{ 0.0275 }\]
\[\approx8444.89 \]