# Thread: [MOVED] $10K initial investment,$1K added annually, at 10%

1. ## [MOVED] $10K initial investment,$1K added annually, at 10%

I am very confused with this problem any help would be appreciated. I know it has to do with some sort of compound interest formula but I been working on this for a week and come up with nothing

Little Mark was given $10,000 at birth from various family members to be invested until his adulthood. The DOW JONES has averaged an annual average growth rate of 10% for the 20th century. If Little Mark invests his$10,000 in a good solid DOW JONES mutual fund which averages 10% growth per year, and then after the first year and every year thereafter ADDS $1000 per year of his own money, how many years will it take for Little Mark to become a millionaire? 2. What annuity (or other) formula did they give you? Thank you! Eliz. 3. Originally Posted by stapel What annuity (or other) formula did they give you? Thank you! Eliz. Hi this is a formula that was given S=〖R[〗⁡█((1+r/n)^nt-1@r@n)[/code] 4. Originally Posted by luvbuggy71 Hi this is a formula that was given S=〖R[〗⁡█((1+r/n)^nt-1@r@n) I'm sorry, but whatever special characters you tried to use are not displaying for me. Please reply using standard formatting, as explained here or here, or use LaTeX. When you reply, please show how far you have gotten in apply the formula. Thank you! Eliz. 5. You need 2 formulas (A = 10,000 and D = 1,000): A(1 + r)^n and D[((1 + r)^n -1)/ r] The initial$10,000 accumulates for n years: 10000(1.10^n)
The annual deposit: 1000[(1.10^n - 1) / .10]

10000(1.10^n) + 1000[(1.10^n - 1) / .10] = 1000000 ; multiply by .10:

1000(1.10^n) + 1000(1.10^n - 1) = 100000 ; divide by 1000:

1.10^n + 1.10^n - 1 = 100

2(1.10^n) = 101

1.10^n = 101/2

n = log(101/2) / log(1.10) = 41.1495.... so a bit over 41 years.

Are you ok with that?

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