Sequences and series questions (1 Viewer)

[Schniz]

helen sets up a prize fund with a single investment of $1000 to provide her school with an annual prize valued at $72. the fund accures interest at a rate of 6% p.a, compounded anually. the first prize is awarded one year after the investment is set up.

(a) Let $B(n) be the balance in the fund at the end of "n" years (and after the nth prize has been awarded). show that B(n) = 1200-200 x (1.06)^n.

(b) at the end of the 10th yr. it is decided to increase the prize value to $90. for how many more years can the prize fund be used to award the prize?

 

[SoulSearcher]

I'm pretty sure this question was asked in the HSC exam a few years back, but I'll answer it for you anyway.

(a)
B₁ = 1000 * 1.06 - 72
B₂ = (1000 * 1.06 - 72) * 1.06 - 72
B₂ = 1000(1.06)² - 72(1.06 + 1)
B₃ = [ 1000(1.06)² - 72(1.06 + 1) ] * 1.06 - 72
B₃ = 1000(1.06)³ - 72(1.06² + 1.06 + 1)
therefore
B_n = 1000(1.06)ⁿ - 72(1.06^n-1 + 1.06^n-2 + ... + 1.06² + 1.06 + 1)
B_n = 1000(1.06)ⁿ - 72 * [ a(rⁿ -1) / (r - 1) ]
B_n = 1000(1.06)ⁿ - 72 * [ (1.06ⁿ -1) / (1.06 - 1) ]
where a =1, r = 1.06
B_n = 1000(1.06)ⁿ - 72 * [ 1.06ⁿ - 1 / 0.06 ]
B_n = 1000(1.06)ⁿ - 1200(1.06ⁿ - 1)
therefore by expanding,
B_n = 1200 - 200 * 1.06ⁿ

(b)Balance at the end of the 10th year
B₁₀ = 1200 - 200 * 1.06¹⁰
B₁₀ = $841.83
After the next year,
B₁₁ = $841.83(1.06) - 90
and thus
B₁₂ = [ $841.83(1.06) - 90 ] * 1.06 - 90
B₁₂ = $841.83(1.06)² - 90(1.06 + 1)
therefore
B_{10 + n} = $841.83(1.06)ⁿ - 90(1.06^n-1 1.06^n-2 + ... + 1.06 + 1)
B_{10 + n} = $841.83(1.06)ⁿ - 90 [ 1.06ⁿ - 1 / 0.06 ]
B_{10 + n} = $841.83(1.06)ⁿ - 1500(1.06ⁿ - 1)
therefore
B_{10 + n} = 1500 - 658.17(1.06)ⁿ

let B_{10 + n} = 0

0 = 1500 - 658.17(1.06)ⁿ
1.06ⁿ = 1500 / 658.17
n ln 1.06 = ln 1500 / 658.17
n = ln (1500 / 658.17) / ln 1.06
n = 14 to nearest whole number
therefore the prize can be awarded for 14 more years

 

[Schniz]

thnx