Hello!
a) It's a geometric series, a = 15000 and r=0.8
so after 2 years, the machine is worth 15000 x 0.8^2 = 9600
b) In the n-th year, the machine is worth ar^(n-1) < 500
15000 x 0.8^(n-1) < 500
0.8^(n-1) < 1/30
(n-1) log 0.8 < log (1/30)
n-1 < (log 1/3)/(log 0.8)
n-1 < 15.24 so after 16 years.
(Check: 15000 x 0.8^16 = 422.21 )
c) Year 1 start: 1000
Year 1 end: 1.05 x 1000
Year 2 start: 1000 + 1.05x1000
Year 2 end: 1.05x(1000+ 1.05 x 1000) = 1.05x1000 + 1.05^2 x 1000
Year 3 start: 1000+ 1.05x 1000 + 1.05^2 x 1000
so year 5 start (say) =
1000 + 1.05x1000 + 1.05^2 x 1000 + 1.05^3 x 1000 + 1.05^4 x 1000
= 1000 (1 + 1.05 + 1.05^2 + 1.05^3 + 1.05^4)
this is a geometric series, a=1, r=1.05
so after 16 years,
amount is worth
1000 [sum to 16 terms of that geometric series]
= 1000 [1(1 - 1.05^16)/(1-1/05)]
= 1000 (21.57856)
= 21578.56
love danniella