ATAR Notes: Forum
HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Advanced => Topic started by: Morrice on December 01, 2018, 08:51:45 pm
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Solution seems wrong, so I'm curious to what the answer may be.
I got $42 509.53
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Yeah, their solution has a mistake. However your's has a minor error as well.
You wrote that \( S_{18} = 1000\left( \frac{1.02^{4\times 19}-1}{1.02-1} \right) \), but it should in fact be \( S_{18} = 1000 \left( \frac{1.02^{4\times 19}-1}{1.02^4-1} \right) \) - the common ratio is \(1.02^4\) so the denominator also has that ^4 in it.
Regardless, thanks for bringing this up. You're most certainly correct in that there should be nineteen terms in the sum, instead of 18. I will forward a message to my manager.
Edit: You're also right in that it should be 1000, not 500.
Edit #2: Upon further glance, it looks like your final answer was correct. I think you knew what you were doing and just didn't write it down correctly but no biggie - message is passed on.
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Yeah, their solution has a mistake. However your's has a minor error as well.
You wrote that \( S_{18} = 1000\left( \frac{1.02^{4\times 19}-1}{1.02-1} \right) \), but it should in fact be \( S_{18} = 1000 \left( \frac{1.02^{4\times 19}-1}{1.02^4-1} \right) \) - the common ratio is \(1.02^4\) so the denominator also has that ^4 in it.
Regardless, thanks for bringing this up. You're most certainly correct in that there should be nineteen terms in the sum, instead of 18. I will forward a message to my manager.
Edit: You're also right in that it should be 1000, not 500.
Edit #2: Upon further glance, it looks like your final answer was correct. I think you knew what you were doing and just didn't write it down correctly but no biggie - message is passed on.
Thank you for reviewing. Yes I approached the problem from a different angle and was lazily writing parts of it next to the solution for contrast, only to miss out on the ^4.
While we're at it, there's also a slight error in the solution to the very first question of the same test, if you don't mind checking it out.
Thanks again.