Question

$2000 is deposited into an account for n years at an interest rate of 16% per annum compounded quarterly. After two years the interest rate changes to 13% per annum compounded monthly. At the end of the term the value of the investment is $4200. Calculate the value of n.

Answer 1

compound interest formula,

$final\ value=P*(1+i)^k\\$

where, i=interest rate k=number of periods.

compounded quarterly for first two years. Therefore i and k are consider quarterly,

For first two years,

$i=4\%\\k=8 \\p=\$2000$

F₁ is the final value after two years,

$F_1=2000(1+0.04)^8$

compounded monthly for next n-2 years. Therefore i and k are consider monthly

For next n-2 years,

$i=\frac{13}{12}\% \\ k=12(n-2)\\p=F_1\\final\ value=\$4200$

$4200=F_1(1+\frac{13}{12}*0.01)^{k}\\ (1+\frac{13}{12}*0.01)^{k}=\frac{4200}{F_1}\\ k*\ln(1+\frac{13}{12}*0.01)=ln(\frac{4200}{2000(1+0.04)^8})\\ k=39.74$

$k=12(n-2)\\ n=\frac{k}{12}+2=\frac{39.74}{12}+2=5.312$

n=5.312years=5 years and 4 months.