super annuation again (1 Viewer)

[bachviete]

Hi, I have done a few particular questions and while thinking about it i dont seam to see how the question can be correctly answered. I would love for some input towards this issue. I will just give an example. "Xuan is saving up for a holiday. She invests $800 at the ed of each year at 7.5% p.a. How much will she have for her holiday after 5 years time." I though of it as that its the start of the year and she puts in $800 at the end of the year, hence A₁ = 800, then a year later she puts in A₂ = A₁ x 1.075 + 800 = 800 x 1.075 + 800. Etc And so at the end of the 4th year, which will be the start of the 5th year and so you just do the geometric series as usual. But after looking at the solution i concluded that they started the period when she first puts the money in the bank. So my question is this, what do we assume when we get questions like this, do we always assume that the starting period is when they first put the first amount of money in the bank? Thanks every one for the help =)

 

[smelnizzle]

This question is actually quite easy =] It shouldn't be too hard to understand lol.
When I see these kinds of questions, I like to create a formula for how much money will be the account after n years. To do this, you need to assess the information given.
Let A1 be the amount of money in the account after 1 year. As the question says "at the end of every year" A1 = 5000
A2 (The amount after the 2nd year) will have the interest applied to it so:
A2 = A1(1 + 6/100) + 5000 (The extra 5000 is because they've deposited another 5000 at the end of the second year.
A2 = 5000(1 + 6/100) + 5000
Similarly for A3:
A3 = A2(1 + 6/100) + 5000
Expanding gives:
A3 = 5000(1 + 6/100)^2 + 5000(1 + 6/100) + 5000
Hopefully by now, a pattern should start to emerge. It looks as if the first term in the sum is always one less power than the year, and the next term is one less than that, and the next one is one less than that until you get to a power of 1.
So:
An = 5000(1 + 6/100)^(n - 1) + 5000(1 + 6/100)^(n - 2) ... + 5000^1
If you take the common factor of 5000 out, you get
An = 5000[1 + 6/100)^(n - 1) + (1 + 6/100)^(n - 2) + ... + 1]
If you reverse the order of the terms in the brackets, to make the 1 first, hopefully you should notice the sum is taking on the form of a geometric series where a=1 r=(1 + 6/100). First though, substitute for n=10.
A10 = 5000[1 + (1 + 6/100) + ... + (1 + 6/100)^9]
Notice how the brackets have 10 terms even though the highest power is 9.
Use the geometric sum formula where r>1:
A10 = 5000[{1(1 + 6/100)^10 - 1}/{(1 + 6/100) - 1}]
And if all is done correctly, you should get an answer of $65903.97 which makes sense, because if 5000 dollars is donated for 10 years, the answer has to be at least 50000.

I hope I haven't made any mistakes lol. Still, the process is the same every time.
Good luck for Tuesday!

hope that helps! when i read this.. i realised what i said prior was wrong, my bad! D: