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Sequence Series Problem (1 Viewer)

annabackwards

<3 Prophet 9
Joined
Jun 14, 2008
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4,670
Location
Sydney
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HSC
2009
For part i
A = P (1 + r/100)^n where P= 10 000, r = 8, n = 10.
= 21 589.25 by calculator (just sub in the numbers)

For part ii
Amount on 1 July 1986
=10 000(1.08) + 1000

Amount on 1 July1987
= [10 000(1.08) + 1000]1.08 +1000
= 10 000(1.08)^2 + 1000(1.08)+ 1000

Amount on 1 July 1988
= [10 000(1.08^)2 + 1000(1.08) + 1000]1.08 + 1000
= 10 000(1.08)^3 + 1000(1.08)^2 + 1000(1.08) + 1000

Amount on 1 July 1995
=10 000(1.08)^10 + 1000(1.08)^9 + 1000(1.08)^8 + ... +1000
=10 000(1.08)^10 + 1000(1 + 1.08 + 1.08^2 +...+ 1.08^9)
=10 000(1.08)^10 + 1000[1(1.08^10 - 1)/(1.08 - 1)] (using the formula for GP sum)
=21 589.25 + 14 486.56 (calculator)
=$36 075.81

For part iii
35 478= 10000(1+ r/100)^10
3.5478 = (1 + r/100)^10
1 + r/100 = (3.5478)^1/10
1 + r/100 = 1.1350 (bycalculator)
r/100 = 0.1350
r = 13.5% (approx.)

I sort of skipped the 1st part of part ii but it was way too long and I tired.
I recommend that you buy the Success One HSC books for maths as they have the actual HSC exam and worked solutions... very useful when I was studying for my 2U HSC :)
 

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