Question

If $3000 is invested at 3% interest, find the value of the investment at the end of 7 years if the interest is compounded as follows. (Round your answers to the nearest cent.)
(i) annually
(ii) semiannually
(iii) monthly
(iv) weekly
(v) daily
(vi) continuously

Answer 1

Answer:

annualy=$3689.62

semiannually=$3695.27

monthly=$3700.06

weekly=$3700.81

daily=$3701.00

Continuously=$3701.03

Step-by-step explanation:

Given:

P=3000

r=3%

t=7 years

Formula used:

Where,

A represents Accumulated amount

P represents (or) invested amount

r represents interest rate

t represents time in years

n represents accumulated or compounded number of times per year

Solution:

(i)annually

n=1 time per year

$A=3000[1+\frac{0.03}{1} ]^1^(^7^)\\ =3000(1.03)^7\\ =3689.621596$

On approximating the values,

A=$3689.62

(ii)semiannually

n=2 times per year

$A=3000[1+\frac{0.03}{2}^{2(4)} ]\\ =3000[1+0.815]^14\\ =3695.267192$

On approximating the values,

A=$3695.27

(iii)monthly

n=12 times per year

$A=3000[1+\frac{0.03}{12}^{12(7)} \\ =3000[1+0.0025]^84\\ =3700.0644$

On approximating,

A=$3700.06

(iv) weekly

n=52 times per year

$A=3000[1+\frac{0.03}{52}]^3^6 \\ =3000(1.23360336)\\ =3700.81003$

On approximating,

A=$3700.81

(v) daily

n=365 time per year

$A=3000[1+\frac{0.03}{365}]^{365(7)} \\ =3000[1.000082192]^{2555}\\ =3701.002234$

On approximating the values,

A=$3701.00

(vi) Continuously

$A=Pe^r^t\\ =3000e^{\frac{0.03}{1}(7) }\\ =3000e^{0.21} \\ =3000(1.23367806)\\ =3701.03418$

On approximating the value,

A=$3701.03