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Geometric Sequence (P2)
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Aleksander Krol
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#1
Q. A virus is spreading such that the number of people infected increases by 4% a day.
Initially 100 people were diagnosed with the virus.
How many days it be before 1000 people are infected?
My working:
It's a geometric sequence,
•so (common ration) r=(100+4)/100 = 1.04
•a=100
•the sequence,
=> 1st day: 100
=> 2nd day: 100(1.04) = 104
=> 3rd day: 104(1.04) = 108.16
•so nth term= ar^(n-1) ,
[where 'a' is the first term of the sequence, 'r' is the common ratio of the sequence, and 'n' is the nth term of the sequence.]
=》nth term= 1000, then, 1000= 100(1.04)^(n-1)
log10 = (n-1) log1.04
n= [ (log10)/(log1.04) ] +1 =59.7
which means, after 59 days, on 60th day, at some time before midnight, exactly 1000 people will be infected.
so, final answer would be, 60 days to infect 1000 people.
But in the ms, it's 59 days and for the nth term they considered : ar^(n).
Why not ar^(n-1) ?
Initially 100 people were diagnosed with the virus.
How many days it be before 1000 people are infected?
My working:
It's a geometric sequence,
•so (common ration) r=(100+4)/100 = 1.04
•a=100
•the sequence,
=> 1st day: 100
=> 2nd day: 100(1.04) = 104
=> 3rd day: 104(1.04) = 108.16
•so nth term= ar^(n-1) ,
[where 'a' is the first term of the sequence, 'r' is the common ratio of the sequence, and 'n' is the nth term of the sequence.]
=》nth term= 1000, then, 1000= 100(1.04)^(n-1)
log10 = (n-1) log1.04
n= [ (log10)/(log1.04) ] +1 =59.7
which means, after 59 days, on 60th day, at some time before midnight, exactly 1000 people will be infected.
so, final answer would be, 60 days to infect 1000 people.
But in the ms, it's 59 days and for the nth term they considered : ar^(n).
Why not ar^(n-1) ?
Last edited by Aleksander Krol; 4 months ago
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BlueChicken
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#2
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#2
(Original post by Aleksander Krol)
Q. A virus is spreading such that the number of people infected increases by 4% a day.
Initially 100 people were diagnosed with the virus.
How many days it be before 1000 people are infected?
My working:
It's a geometric sequence,
•so (common ration) r=(100+4)/100 = 1.04
•a=100
•the sequence,
=> 1st day: 100
=> 2nd day: 100(1.04) = 104
=> 3rd day: 104(1.04) = 108.16
•so nth term= ar^(n-1) ,
[where 'a' is the first term of the sequence, 'r' is the common ratio of the sequence, and 'n' is the nth term of the sequence.]
=》nth term= 1000, then, 1000= 100(1.04)^(n-1)
log10 = (n-1) log1.04
n= [ (log10)/(log1.04) ] +1 =59.7
which means, after 59 days, on 60th day, at some time before midnight, exactly 1000 people will be infected.
so, final answer would be, 60 days to infect 1000 people.
But in the ms, it's 59 days and for the nth term they considered : ar^(n).
Why not ar^(n-1) ?
Q. A virus is spreading such that the number of people infected increases by 4% a day.
Initially 100 people were diagnosed with the virus.
How many days it be before 1000 people are infected?
My working:
It's a geometric sequence,
•so (common ration) r=(100+4)/100 = 1.04
•a=100
•the sequence,
=> 1st day: 100
=> 2nd day: 100(1.04) = 104
=> 3rd day: 104(1.04) = 108.16
•so nth term= ar^(n-1) ,
[where 'a' is the first term of the sequence, 'r' is the common ratio of the sequence, and 'n' is the nth term of the sequence.]
=》nth term= 1000, then, 1000= 100(1.04)^(n-1)
log10 = (n-1) log1.04
n= [ (log10)/(log1.04) ] +1 =59.7
which means, after 59 days, on 60th day, at some time before midnight, exactly 1000 people will be infected.
so, final answer would be, 60 days to infect 1000 people.
But in the ms, it's 59 days and for the nth term they considered : ar^(n).
Why not ar^(n-1) ?
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Aleksander Krol
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#3
(Original post by BlueChicken)
It might be the wording of the questions. I would assume 100 is start, so after 1 day number is 100 x 1.04, after 2 days it's 100 x 1.04 x 1.04 etc. Which would be of the form arn.
It might be the wording of the questions. I would assume 100 is start, so after 1 day number is 100 x 1.04, after 2 days it's 100 x 1.04 x 1.04 etc. Which would be of the form arn.
this question is similar to the first question. in the second question, his distance increases 10% a day and in the first question, no. of people infected increases by 4% a day,
then why is it that we use ar^(n-1) for the second question?
Last edited by Aleksander Krol; 4 months ago
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BlueChicken
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#4
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#4
(Original post by Aleksander Krol)
"Q2. Richard is sponsered to cycle 1000 miles over a number of days. he cycles 10 miles on day 1 and increases the distance by 10% a day. How long will it take him to complete the challenge?"
this question is similar to the first question. in the second question, his distance increases 10% a day and in the first question, no. of people infected increases by 4% a day,
then why is it that we use ar^(n-1) for the second question?
"Q2. Richard is sponsered to cycle 1000 miles over a number of days. he cycles 10 miles on day 1 and increases the distance by 10% a day. How long will it take him to complete the challenge?"
this question is similar to the first question. in the second question, his distance increases 10% a day and in the first question, no. of people infected increases by 4% a day,
then why is it that we use ar^(n-1) for the second question?
You need to think what is the starting point and after 1 day what has happened. In first instance, the amount has increased from 100 to 104. In second it is 10 (he started at 0).
Last edited by BlueChicken; 4 months ago
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