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# C2 geometric q help. Watch

1. The sum to infinity is double the first term. Find the common ratio.

How do I go about it ??

Thanks
2. Start with 2a = a/(1-r) and see if you can work it out from there
3. (Original post by Kieran?)
Start with 2a = a/(1-r) and see if you can work it out from there
Thanks, i had that, but i solved it wrongly,
4. (Original post by Hi, How are you ?)
Thanks, i had that, but i solved it wrongly,
Divide by a and you get 2 = 1/(1-r)

From there you just simply rearrange it to get r
5. (Original post by Kieran?)
Divide by a and you get 2 = 1/(1-r)

From there you just simply rearrange it to get r
can you help with 1 more q please, thanks,

There are 400 deers in jan 08
the population grows by 10% per year, but 30 deer are culled at the end of each year.

How many deers will there be in jan 19?
6. (Original post by Hi, How are you ?)
can you help with 1 more q please, thanks,

There are 400 deers in jan 08
the population grows by 10% per year, but 30 deer are culled at the end of each year.

How many deers will there be in jan 19?
a=400, r=1.1 and n=11

Apply the normal sum for a geometric series with those values

After that you need to account for the 30 lost at the end of each year. So 30 x 11 = 330 lost over the 11 years. Then you simply take 330 away from the value you got for your previous sum
7. (Original post by Kieran?)
a=400, r=1.1 and n=11

Apply the normal sum for a geometric series with those values

After that you need to account for the 30 lost at the end of each year. So 30 x 11 = 330 lost over the 11 years. Then you simply take 330 away from the value you got for your previous sum
its not the sum of the deers, its the amount of individual deers.
8. (Original post by Hi, How are you ?)
its not the sum of the deers, its the amount of individual deers.
Not sure what you mean by that, if you use the Sn = a(1-r^n)/(1-r) formula then it's the same as adding up the number of deer in each year individually...
9. (Original post by Kieran?)
a=400, r=1.1 and n=11

Apply the normal sum for a geometric series with those values

After that you need to account for the 30 lost at the end of each year. So 30 x 11 = 330 lost over the 11 years. Then you simply take 330 away from the value you got for your previous sum
This doesn't work, you cannot take the deer away at the end as this would increase the number comming in each year e.g. if you have 1 and add 10% then minus 0.1 you will always have 1 but with your method you would get 1.01.
10. (Original post by james22)
This doesn't work, you cannot take the deer away at the end as this would increase the number comming in each year e.g. if you have 1 and add 10% then minus 0.1 you will always have 1 but with your method you would get 1.01.
oh yeah... woops
how would you go about it then...?

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