T 4–4 Deposits needed to accumulate a future sum

  • Context: MHB 
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Future Sum
Click For Summary
SUMMARY

Judi aims to accumulate $8,000 in 5 years through equal annual end-of-year deposits, with an interest rate of 7%. The formula used to determine the required deposit amount is derived from the future value of an annuity formula, specifically $D = \frac{A_n i}{(1+i)^n - 1}$. By substituting $A_n = 8000$, $i = 0.07$, and $n = 5$, the necessary annual deposit can be calculated. This approach effectively utilizes the principles of financial mathematics to solve for the deposit amount.

PREREQUISITES
  • Understanding of future value of annuities
  • Familiarity with basic algebra and equations
  • Knowledge of interest rates and their application in finance
  • Ability to manipulate exponential equations
NEXT STEPS
  • Calculate the annual deposit required using the formula $D = \frac{8000 \times 0.07}{(1 + 0.07)^5 - 1}$
  • Explore the concept of annuities and their types in financial planning
  • Learn about different investment strategies to maximize returns
  • Study the impact of varying interest rates on future savings
USEFUL FOR

Individuals planning for future savings, financial analysts, students studying finance, and anyone interested in understanding the mechanics of annuities and investment growth.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
T 4–4 Deposits needed to accumulate a future sum Judi wishes to accumulate \$8,000 by the end of 5 years by making equal annual end-of-year deposits over the next 5 years. If Judi can earn 7% on her investments, how much must she deposit at the end of each year to meet this goal?

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

ok not sure how plug this in

this complicated by the deposit made at the end of each year
 
Last edited:
Physics news on Phys.org
Let's let $D$ be the amount of the end of year deposits, and $i$ be the annual interest rate. So, the amount of the account at the end of year $n$ can be given by the difference equation:

$$A_n-(1+i)A_{n-1}=D$$ where $n\in\mathbb{N}$

The homogeneous solution is given by:

$$h_n=k_1(1+i)^n$$

And the particular solution is:

$$p_n=k_2$$

Plugging this into our difference equation, we find:

$$k_2-(1+i)k_2=D\implies k_2=-\frac{D}{i}$$

And so the closed form for $A_n$ is given by:

$$A_n=k_1(1+i)^n-\frac{D}{i}$$

Since:

$$A_1=D$$

We find:

$$k_1=\frac{D}{i}$$

And so the closed-form for $A_n$ is

$$A_n=\frac{D}{i}\left((1+i)^n-1\right)$$

Solve for $D$:

$$D=\frac{A_ni}{(1+i)^n-1}$$
 
good grief,,😎
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
3
Views
2K
Replies
1
Views
7K
  • · Replies 2 ·
Replies
2
Views
7K
Replies
1
Views
3K
  • · Replies 27 ·
Replies
27
Views
4K
Replies
10
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K