Question

The sum of the first nth term of a geometric series is 145 and the sum of the reciprocal is 145/33. The first term is 1. What is n and the common ratio?

Answer 1

There is something wrong with the problem statement because the solution does not yield an integer value of n.

Explanation:

Using this Geometric Series reference we obtain the following equation:

`sum_(k=0)^(n-1) ar^k= a(1-r^n)/(1-r)`

The first two sentences of the problem statement tell us that the following 2 equations are true:

`(1-r^n)/(1-r) = 145" [1]"`

`(1-r^-n)/(1-r^-1)= 145/33" [2]"`

Divide equation [1] by equation [2]:

`(1-r^n)/(1-r)(1-r^-1)/(1-r^-n)= 145 33/145`

Simplify the right side and multiply the left side by 1 on the form of `r^n/r^n`:

`(1-r^n)/(1-r)(r^n-r^(n-1))/(r^n-1)= 33`

Observing that there is a common factor of `(-1)/-1`:

`(1-r^n)/(1-r)(r^(n-1)-r^n)/(1-r^n)= 33`

`(1-r^n)/(1-r^n)` becomes 1:

`(r^(n-1)-r^n)/(1-r)= 33`

Factor the numerator:

`(r^(n-1)(1-r))/(1-r)= 33`

Cancel the common factor:

`r^(n-1)= 33`

Multiply both sides by `r` and designate the equation as [3]:

`r^n= 33r" [3]"`

Substitute equation [3] into [1]:

`(1-33r)/(1-r) = 145" [1.1]"`

`r=9/7`

Substitute `r = 9/7` into equation [1]:

`(1-(9/7)^n)/(1-(9/7)) = 145`

`n ~~ 14.9129`