# Formula for annuity payments

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1. Jan 13, 2016

### NameIsUnique

So, unless you've been living under a rock, you should know the jackpot for the powerball is at 1.5 billion dollars.

I was looking up the distribution of annuity payments and the website said that the payments are not equally distributed. They are incremented by 5% each year.

Like the nerd I am, I tried figuring out the math but didn't know where to start.

I know that 1.5 billion / 30 payments = 50 million a year (before taxes)

How would you go about figuring out 30 payments equating to 1.5 billion but the condition is that each payment is 5% more than the next?

2. Jan 13, 2016

### Krylov

Is the amount increasing or decreasing in time?

If the total amount is $p$ (= 1.5 billon dollars) and the rate of increase / decrease is $\lambda$ (= 1.05 or 0.95) and your amount in the $k$th year is $a_k$, then $a_k = \lambda^{k-1}a_1$ for $k = 1,\ldots,n$ where $n$ is the amount of years. Now set $\sum_{k=1}^n{a_k} = p$ (geometric sum) and solve for $a_1$.

3. Jan 13, 2016

### NameIsUnique

Increasing.

Year 1 = x
Year 2 = x +(x* 0.05)
and keeps incrementing until 1.5 billion

4. Jan 13, 2016

### Krylov

Ok, so then you set $\lambda = 1.05$, take the geometric sum and solve the equation for $a_1$. Once $a_1$ is known, use the formula for $a_k$ to compute the amount in year $k$.

5. Jan 13, 2016

### NameIsUnique

Thanks a lot!

6. Jan 13, 2016

### Krylov

Don't mention it. Just be sure to let me know if you win the jackpot

7. Jan 13, 2016

### NameIsUnique

I just plugged it in year one

I think I'm doing it wrong.

1.5 billion = (1.05) ^ (1-1) * a1

and I solve for a1?

It seems like I get 1.5 billion

8. Jan 13, 2016

### NameIsUnique

Nvm I get it.

9. Jan 13, 2016

### Krylov

No, that is not correct. You need to solve
$$a_1\sum_{k=1}^n{\lambda^{k-1}} = p$$
First you need to evaluate the sum, using the standard formula for the geometric sum. I leave that up to you as a challenge. Once that is done, you can solve for $a_1$.

10. Jan 13, 2016

### Krylov

Ok, very well!