Geometric Progression Question

[crays]

Hi. Tried solving, but no idea how.

At the end of 1995, the population of Ubris was 46650 and by the end of 2000 it had risen to 54200. On the assumption that the population at the end of each year form a g.p. find

a) The population at the end of 2006, leaving your answer in 3 s.f.

b) The year in which the population reaches 100000, correct to nearest integer.

 

[rock.freak667]

This equation will help you

$$T_n=ar^{n-1}$$

in 1995 n=0, in 2000 what does n=?

you can solve now

 

[crays]

but how sure are you that 1995 is the starting year @_@ ?

Well i got the answer correct with that eqn. But can you briefly explain why must i use n = 12 instead of n = 11 for question a ?

 

[Fightfish]

You can arbitrarily set the starting year; it will just be a shift. The GP will hold throughout. i.e. a could well be $$a = br^{k}$$, where b and k are constants. So the GP is simply just transformed to a higher 'starting point'

 

[crays]

Thanks

 

[Fightfish]

You must use n = 12 as at the starting year 1995, n = 1 since its the first term in the G.P. that has been defined.

 

[HallsofIvy]

There is nothing wrong with using n= 0 for 1995 and then, if you are using $T_n= ar^{n-1}$ your first equation is $ar^{-1}= 46650$. Or use n= 1 so you have $ar^0= a= 46650$. If you are using $br^k$, taking k= 0 for 1995 gives $br^0= b= 46650$ and k= 1 $br^1= br= 46650$. You just get different values of a or b, and r. But where did you get "n= 12"? It is only 5 years from "the end of 1995" to "the end of 2000". If your are taking 1995 to be n= 0, 2000 will be n= 5. If you are taking 1995 to be n= 1, 2000 will be n= 6.