Sequences problem

[Leilau]

Scenario A: Lilly decides to invest $400 in 2001. Her interest rate is 5.5% compounded yearly.
Lilly deposits an additional $1200 per year.
Scenario B: Mark decides to invest $3800 in 2001. His interest rate is 2.5% compounded yearly.
Mark deposits an additional $1200 per year.

1.
How much will Lilly have in 2020?

2.
When will Mark have $50,000?

3.
In what year will Lilly have more money than Mark?

 

[Steven G]

What have you tried? Where are you stuck? This is a math help forum. To receive help you must tell us where you are stuck.

 

[Leilau]

What have you tried? Where are you stuck? This is a math help forum. To receive help you must tell us where you are stuck.

I have many answers to this but don’t really know if I’m on the right path. I started with 400(1.055)^19 + 1200(1.055)^19 due to the fact that 1200 is being compounded annually. So the answer I got for the first one was $4425.04 but I also did it on the graphic calculator and I got an answer of 3962 when I included an added 1200 to the recursive equation u(n) = u(n-1)*(1+.055) +1200.

 

[Steven G]

I have many answers to this but don’t really know if I’m on the right path. I started with 400(1.055)^19 + 1200(1.055)^19 due to the fact that 1200 is being compounded annually. So the answer I got for the first one was $4425.04 but I also did it on the graphic calculator and I got an answer of 3962 when I included an added 1200 to the recursive equation u(n) = u(n-1)*(1+.055) +1200.

400(1.055)^19 + 1200(1.055)^19 clearly says that Lilly is investing $400 at 5.5% for 19 years AND investing $1200 at 5.5% for 19 years.
That is NOT true. Only 400 is being invested at 5.5% for 19 years. Also $1200 will be invested for 18 years and another $1200 will be invested for 17 years and another $1200 will be invested for 16 years and another $1200 will be invested for 15 years and .....another $1200 will be invested for 1 year.
Do you see that? Now can you compute the total interest for the 19 years??