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Superannuation (2 Viewers)

Arithela

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A school invests $5000 at the end of each year at 6% p.a. towards a new library. How much will the school have after 10 years?

Answer: $65903.97

Somebody show full working please!
 

lolokay

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it's a series.
You have 5000*(1.06)9 + 5000*(1.06)8 ... + 5000
[they don't put any money in until the end of year 1. This $5000 is compounded over 9 years, the second $5000 compounded over 8 years etc. and then $5000 at the end of the 10th year which is not compounded. I think the question is a bit ambiguous]


= (5000*(1.06)10 - 5000)/0.06
[which you should know from geometric series]

= 65903.97
 

Timothy.Siu

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A school invests $5000 at the end of each year at 6% p.a. towards a new library. How much will the school have after 10 years?

Answer: $65903.97

End of 1st year - $5000
End of 2nd year - $5000x1.06+$5000
End of 3rd year - ($5000x1.06+$5000)x1.06+$5000=$5000x1.06^2+$5000x1.06+$5000
End of nth year - $5000x1.06^(n-1)+$5000x1.06^(n-1)+.....+$5000
=($5000(1.06^n - 1))/1.06-1 (geometric series sum formula)

therefore end of 10th year = ($5000(1.06^10 - 1))/1.06-1

= $65903.97471=$65903.97 (nearest cent)
 

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