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# GP Watch

1. Hi

A girl invests 500 at the start of each year in an account paying 3.5% interest per year. At the start of each year she puts in another 500. After how many years has this account accumulated more than 20000?

Exam solutionz has a different approach but I took a different approach because I didn't understand his approach.

So, i sed the account starts with 500 and the multiplicative factor of 3.5% will be 1.035 so that will be r

so a = 500
r = 1.035

I then used my formula for Summing Gps and ended up with n = log2.4/log1.035
= 25.4486

Exam Solutionz ended up with n being greater than or equal to 24.86...
2. (Original post by NinCheng)
...
Quick skim, how does your method account for the extra 500 she adds in every year?
3. (Original post by Zacken)
Quick skim, how does your method account for the extra 500 she adds in every year?
idk but I got 25.44 was that by luck or
4. (Original post by NinCheng)
idk but I got 25.44 was that by luck or
Could you post a picture of the question or link to the examsolutions video or something?
5. (Original post by Zacken)
Could you post a picture of the question or link to the examsolutions video or something?
http://www.examsolutions.net/maths-r.../example-3.php
6. So your answer isn't correct. You need to write down the first few terms of the sequence and see if you notice something.

1st year: 500(1.05)
2nd year: (500 + 500(1.05))(1.05) = 500(1.05) + 500(1.05)^2
3rd year: 500(1.035) + 500(1.035)^2 + 500(1.035)^3

So nth year: 500(1.035 + 1.035^2 + ... + 1.035^n)

So you want to find n s.t 500(1.0.35 + 1.035^2 + ... + 1.035^n) > 20000

The inside of the bracket is a sum you can use your formula for.

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