# Calculating future value on an initial deposit and then, incrementing this initial deposit as a recurring deposit yearly

I have a scenario where \$10000 was initially invested into a savings account, at 6% pa compounded yearly. An amount of \$1000 is then added annually to this savings amount after the first year, and we are to determine the value of the total investment at the end of 10 years. In my mind, its this initial deposit of \$10000 that gets compounded for 11 years at 6%, then an annuity of \$1000 per year that you calculate over the remaining 10 years, but this is apparently wrong.

• No, that's correct - an initial deposit and a 10-year annuity. What formulas are you using for each? Aug 17, 2022 at 18:49
• At least from your paraphrase of your task, I would assume it's 10 years (not 11) for the initial deposit and a 9 year annuity (or maybe 10 years with end-of-year deposit, depending on wether you are supposed to add the last deposit or not). Do you have the correct result? Aug 18, 2022 at 9:33

You treat them indeed as two independent investment

1. 10,000 at 6% for 10 years -> \$17,908.48
2. 1,000/year at 6% for 9 years -> \$12,180.79

The trickiest part is probably to decipher from the original text what the exact timing is and to get the number of periods correctly.

With annual interest `r = 0.06`

Similarly to Hilmar's answer, compound the 10000 and add the annuity, but with `n = 10`

``````10000 (1 + r)^11 + (1000 (1 + r) ((1 + r)^n - 1))/r =
18982.99         + 13971.64                         =
32954.63
``````

The `n = 10` means 10 cash flows of 1000, with the last deposit gaining one year's interest.

In the annuity formula `d` is the regular deposit.

$\sum_{k=1}^{n}d(1+r)^k=\frac{d(1+r)((1+r)^n-1)}{r}$

To be clear on the balance of cash flows, 32954.63 can be withdrawn at the end of year 11.

(End of year 0 is the start of year 1.)

``````Yr end  Cash flow
0       10000
1        1000
2        1000
3        1000
4        1000
5        1000
6        1000
7        1000
8        1000
9        1000
10       1000
11     -(18982.99 + 13971.64) = -32954.63
``````