# Depreciation and Value

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• Published : October 5, 2011

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i. After knee surgery, your trainer tells you to return to your jogging program slowly. He suggests jogging for 12 minute each day for the first week. Each week thereafter, he suggests that you increase that time by 6 minutes per day. How many weeks will it be before you are up to jogging 60 minutes per day? Solution:

(12, 18, 24, 30…… 60)
T1 = 12, T2 = 18, T3=24, T4 = 30, Tn = 60
T2 – T1 = T3 – T2
18 – 12 = 24 – 18
6 = 6
This above is arithemetic sequence.
Common difference (d) = 6
First term (a) = 12
Tn = a + (n – 1) d
60 = 12 + (n – 1) 6
60 = 12 + 6n – 6
60 = 6n + 6
n = 9
*The week that I can jog 60 minutes in a day is ninth weeks. Therefore, it would be eight weeks before I am up to jogging 60 minutes per day.

ii. A culture of bacteria doubles every 2 hours. If there are 500 bacteria at the beginning, how many bacteria will there be after 24 hours? Solution:
500, 1000, 2000, 4000……
Start 2hours, 4hours, 6hours
T1 = 2, T 2 = 4, T3=6 T1 = 500, T2 = 1000, T3= 2000, T4 = 4000 T2 – T1 = T3 – T2 T2T1=T3T2
4 – 2 = 6 – 4 1000500=20001000
2 = 2 2 = 2
This above is arithemetic sequence. This above is geometric sequence. Common difference (d) = 2 Common ratio (r) = 2
First term (a) = 2 First term (a) = 500 Tn = 24 Tn = 12
Tn = a + (n – 1) d Tn = arn-1
24 = 2 + (n – 1) 2 T12 = 500(2)12-1 24 = 2 + 2n – 2 T12 = 500(2)11
2n = 24 – 2 + 2 T12 = 500 × 2048
2n = 24 T12 = 1024 000
n = 12

Therefore, if there are 500 bacteria at the beginning, 1024 000 bacteria will there be after 24 hours.