Question

Suppose that an account was opened with a $1,437 deposit at the end of 2091 and that there were no additional deposits. The nominal interest rate is 11.7%; compounding occurs 11 times per year. What is the minimum number of periods required to accumulate $3,117? (answer: 74)

Please show all work on how you arrived at that answer. Thanks

Answer 1

We use the formula:
A=P(1+r/11)^n
where
A=future value
P=present value
r=rate of interest
n=time period.

3117=1437*(1+0.117/11)^n

(3117/1437)=(1+0.117/11)^n

Taking log on both sides;

log (3117/1437)=n*log (1+0.117/11)

n=log (3117/1437)/log (1+0.117/11)

n=[log (3117/1437)/log (1+0.117/11)]

which is equal to

=74 periods. (Approx)