If 1700 dollars is invested in an account for 10 years. Find the value of the investment at the end of 10 years if the interest is:

(a) 4.2% compounded annually: $ 

(b) 4.2% compounded semiannually: $ 

(c) 4.2% compounded quarterly: $ 

(d) 4.2% compounded monthly: $ 

(e) 4.2% compounded daily (ignore leap years): $ 


Round answers to the nearest cent.

To solve these problems, we will use the formula for compound interest:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $A$ is the future value of the investment.
- $P$ is the principal amount (initial investment), which is $1700.
- $r$ is the annual interest rate (as a decimal), which is 4.2% or 0.042.
- $n$ is the number of times the interest is compounded per year.
- $t$ is the time the money is invested for, which is 10 years.

### Part (a): 4.2% compounded annually

Here, $n = 1$ because the interest is compounded annually.

\[
A = 1700 \left(1 + \frac{0.042}{1}\right)^{1 \times 10}
\]
\[
A = 1700 \left(1.042\right)^{10}
\]
\[
A = 1700 \times 1.519406
\]
\[
A \approx 2582.99
\]

So, the value of the investment after 10 years compounded annually is **$2582.99**.

### Part (b): 4.2% compounded semiannually

Here, $n = 2$ because the interest is compounded semiannually.

\[
A = 1700 \left(1 + \frac{0.042}{2}\right)^{2 \times 10}
\]
\[
A = 1700 \left(1.021\right)^{20}
\]
\[
A = 1700 \times 1.521640
\]
\[
A \approx 2586.79
\]

So, the value of the investment after 10 years compounded semiannually is **$2586.79**.

### Part (c): 4.2% compounded quarterly

Here, $n = 4$ because the interest is compounded quarterly.

\[
A = 1700 \left(1 + \frac{0.042}{4}\right)^{4 \times 10}
\]
\[
A = 1700 \left(1.0105\right)^{40}
\]
\[
A = 1700 \times 1.522812
\]
\[
A \approx 2588.78
\]

So, the value of the investment after 10 years compounded quarterly is **$2588.78**.

### Part (d): 4.2% compounded monthly

Here, $n = 12$ because the interest is compounded monthly.

\[
A = 1700 \left(1 + \frac{0.042}{12}\right)^{12 \times 10}
\]
\[
A = 1700 \left(1.0035\right)^{120}
\]
\[
A = 1700 \times 1.523531
\]
\[
A \approx 2589.50
\]

So, the value of the investment after 10 years compounded monthly is **$2589.50**.

### Part (e): 4.2% compounded daily

Here, $n = 365$ because the interest is compounded daily (ignoring leap years).

\[
A = 1700 \left(1 + \frac{0.042}{365}\right)^{365 \times 10}
\]
\[
A = 1700 \left(1.000115\right)^{3650}
\]
\[
A = 1700 \times 1.523867
\]
\[
A \approx 2589.92
\]

So, the value of the investment after 10 years compounded daily is **$2589.92**.

---

Would you like further details on any of these calculations or have any questions?

Here are five related questions:
1. How does changing the interest rate affect the final amount?
2. What happens if the investment is compounded continuously?
3. How does the duration of the investment affect the final value?
4. What is the difference between simple interest and compound interest?
5. How would an additional yearly contribution affect the outcome?

**Tip**: The more frequently interest is compounded, the greater the final investment value will be.

If 1700 dollars is invested in an account for 10 years. Find the value of the investment at the end of 10 years if the interest is: (a) 4.2% compounded annually: $ (b) 4.2% compounded semiannually: $ (c) 4.2% compounded quarterly: $ (d) 4.2% compounded monthly: $ (e) 4.2% compounded daily (ignore leap years): $ Round answers to the nearest cent.

Find the value of a $1700 investment over 10 years with 4.2% interest compounded annually, semiannually, quarterly, monthly, and daily. Detailed breakdown using the compound interest formula A = P(1 + r/n)^(nt) with results rounded to the nearest cent. Suitable for students in Grades 9-12.

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