Answer :
To determine the annual percentage interest rate required for a lump sum to double in 6 years when the interest is compounded quarterly, we can follow these steps:
1. Understand the Problem:
- We need the final amount to be double the initial amount.
- The investment period is 6 years.
- Interest is compounded quarterly, meaning there are 4 compounding periods per year.
2. Set Up Variables:
- Let [tex]\( P \)[/tex] be the principal amount (initial investment).
- After 6 years, we want the final amount [tex]\( A \)[/tex] to be [tex]\( 2P \)[/tex] (double the initial investment).
- Let [tex]\( r \)[/tex] be the annual interest rate (which we need to find).
- The number of compounding periods per year, [tex]\( n \)[/tex], is 4 (since it's compounded quarterly).
- The total number of compounding periods over 6 years, [tex]\( nt \)[/tex], is [tex]\( 4 \times 6 = 24 \)[/tex].
3. Use the Compound Interest Formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
4. Substitute Known Values:
[tex]\[ 2P = P \left(1 + \frac{r}{4}\right)^{24} \][/tex]
5. Simplify the Equation:
[tex]\[ 2 = \left(1 + \frac{r}{4}\right)^{24} \][/tex]
6. Solve for [tex]\( r \)[/tex]:
- Start by taking the natural logarithm (ln) on both sides of the equation to make it easier to solve for [tex]\( r \)[/tex].
[tex]\[ \ln(2) = \ln\left[\left(1 + \frac{r}{4}\right)^{24}\right] \][/tex]
- Using the property of logarithms [tex]\( \ln(a^b) = b \ln(a) \)[/tex], we get:
[tex]\[ \ln(2) = 24 \ln\left(1 + \frac{r}{4}\right) \][/tex]
- Solving for [tex]\( \ln\left(1 + \frac{r}{4}\right) \)[/tex]:
[tex]\[ \ln\left(1 + \frac{r}{4}\right) = \frac{\ln(2)}{24} \][/tex]
- Taking the exponential on both sides to remove the natural logarithm:
[tex]\[ 1 + \frac{r}{4} = e^{\frac{\ln(2)}{24}} \][/tex]
- Isolate [tex]\( r \)[/tex]:
[tex]\[ \frac{r}{4} = e^{\frac{\ln(2)}{24}} - 1 \][/tex]
[tex]\[ r = 4 \left(e^{\frac{\ln(2)}{24}} - 1\right) \][/tex]
Given the derived formula and assuming the correct computations, the annual interest rate [tex]\( r \)[/tex] that will double the lump sum in 6 years when compounded quarterly is approximately 11.72%.
1. Understand the Problem:
- We need the final amount to be double the initial amount.
- The investment period is 6 years.
- Interest is compounded quarterly, meaning there are 4 compounding periods per year.
2. Set Up Variables:
- Let [tex]\( P \)[/tex] be the principal amount (initial investment).
- After 6 years, we want the final amount [tex]\( A \)[/tex] to be [tex]\( 2P \)[/tex] (double the initial investment).
- Let [tex]\( r \)[/tex] be the annual interest rate (which we need to find).
- The number of compounding periods per year, [tex]\( n \)[/tex], is 4 (since it's compounded quarterly).
- The total number of compounding periods over 6 years, [tex]\( nt \)[/tex], is [tex]\( 4 \times 6 = 24 \)[/tex].
3. Use the Compound Interest Formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
4. Substitute Known Values:
[tex]\[ 2P = P \left(1 + \frac{r}{4}\right)^{24} \][/tex]
5. Simplify the Equation:
[tex]\[ 2 = \left(1 + \frac{r}{4}\right)^{24} \][/tex]
6. Solve for [tex]\( r \)[/tex]:
- Start by taking the natural logarithm (ln) on both sides of the equation to make it easier to solve for [tex]\( r \)[/tex].
[tex]\[ \ln(2) = \ln\left[\left(1 + \frac{r}{4}\right)^{24}\right] \][/tex]
- Using the property of logarithms [tex]\( \ln(a^b) = b \ln(a) \)[/tex], we get:
[tex]\[ \ln(2) = 24 \ln\left(1 + \frac{r}{4}\right) \][/tex]
- Solving for [tex]\( \ln\left(1 + \frac{r}{4}\right) \)[/tex]:
[tex]\[ \ln\left(1 + \frac{r}{4}\right) = \frac{\ln(2)}{24} \][/tex]
- Taking the exponential on both sides to remove the natural logarithm:
[tex]\[ 1 + \frac{r}{4} = e^{\frac{\ln(2)}{24}} \][/tex]
- Isolate [tex]\( r \)[/tex]:
[tex]\[ \frac{r}{4} = e^{\frac{\ln(2)}{24}} - 1 \][/tex]
[tex]\[ r = 4 \left(e^{\frac{\ln(2)}{24}} - 1\right) \][/tex]
Given the derived formula and assuming the correct computations, the annual interest rate [tex]\( r \)[/tex] that will double the lump sum in 6 years when compounded quarterly is approximately 11.72%.