Math Help - geometric series problem

1. geometric series problem

In a city in year 1990 have lived 700000 residents while in year 2000 there were 900000 residents

If you take the growth in population made at the same rate , find :

a) how much is the population grow of the city in the time period 1990-2000 ?

b) how much became the population number this year(2009) ?

c) when will the city have 1400000?

2. Originally Posted by beq!x
In a city in year 1990 have lived 700000 residents while in year 2000 there were 900000 residents

If you take the growth in population made at the same rate , find :
What sort of growth are you supposed to assume? Linear? Exponential? Something else?

When you reply, please include a clear listing of your steps and reasoning so far. Thank you!

3. Originally Posted by beq!x
In a city in year 1990 have lived 700000 residents while in year 2000 there were 900000 residents

If you take the growth in population made at the same rate , find :

a) how much is the population grow of the city in the time period 1990-2000 ?

b) how much became the population number this year(2009) ?

c) when will the city have 1400000?

If the title is assumed to be correct then you can use the general formula for a geometric sequence

Using the sum of a geometric sequence:

$S_n = \frac{a(r^n-1)}{r-1}$

If you set 1990 as n=0 then in 2000: n = 2000-1990. Setting 1990 as 0 will also give you a value of a from which the common ratio r can be found

4. this is the original problem i didn't change anything.
i think at first we have to find $q$.

at c) is it after 27 years ? am i right ?