MHB T 4–4 Deposits needed to accumulate a future sum

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Future Sum
AI Thread Summary
To accumulate $8,000 in 5 years with annual deposits at a 7% interest rate, the formula for future value with regular deposits is applied. The difference equation for the account balance is established, leading to a closed-form solution for the amount after n years. The final formula for the annual deposit, D, is derived as D = (A_n * i) / ((1 + i)^n - 1). By substituting the values, one can calculate the required annual deposit. This approach effectively determines the necessary contributions to reach the savings goal.
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:
Mathematics 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,,😎
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top