Calculus
posted by Anonymous .
A tree grows 2.80 m during the first year since it was planted. During each subsequent year the tree's growth is 85% of its growth the previous year.
a) Calculate to the nearest 0.001m, the growth of the tree in the fourth year.
Tn = ar^n1
T4 = 2.80(0.85)^101
T4 = 2450
I don't get how to do this, the answer is wrong, its supposed to be 1.72m :\
Determine the first year in which the growth of the tree is less than half a metre.

Where in the world did you get 10 in the 101
should have been 41
T4 = 2.8(.85)^3 = 1.719 or 1.72
for growth to be .5
.5 = 2.8(.85)^(n1)
divide by 2.8
.17857.. = .85^(n1)
take log of both sides
log (.17857..) = log [.85^(n1)]
log (.17857..) = (n1)log [.85]
n1 = log (.17857..) / log [.85] = 10.6
n = 11.6
in year 11, growth = 2.8(.85)^10 = .5512
in year 12, growth = 2.8(.85)^11 = .468
so what do you think? 
So that would be Year 12.

yes
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