1. ## Geometric progression

Question:

A geometric progression has first term a (where a > 0), and common ratio r. The sum of the first n terms is Sn and the sum to infinity is S. Given that S2 is twice the value of the fifth term, find the value of r. Hence find the least value of n such that Sn is within 5 % of S.

Thank you for helping!

2. Originally Posted by Tangera
Question:

A geometric progression has first term a (where a > 0), and common ratio r. The sum of the first n terms is Sn and the sum to infinity is S. Given that S2 is twice the value of the fifth term, find the value of r. Hence find the least value of n such that Sn is within 5 % of S.

Thank you for helping!
You want $\displaystyle S_2 = 2*s_5$

where $\displaystyle S_n =\sum_0^n ar^n$, and $\displaystyle s_n = ar^n$

$\displaystyle \sum_0^2 ar^n=2ar^5$

$\displaystyle \sum_0^2 r^n=2r^5$

$\displaystyle 1+r+r^2=2r^5$

$\displaystyle 0=2r^5-r^2-r-1$

3. Hello, Tangera!

Are you sure of the wording of the problem?
It doesn't seem to be solvable.

A geometric progression has first term $\displaystyle a > 0$, and common ratio $\displaystyle r.$
The sum of the first $\displaystyle n$ terms is $\displaystyle S_n$ and the sum to infinity is $\displaystyle S.$

Given that $\displaystyle S_2$ is twice the value of the fifth term, find the value of $\displaystyle r.$
Hence find the least value of $\displaystyle n$ such that $\displaystyle S_n$ is within 5% of $\displaystyle S.$

$\displaystyle S_2$ is the sum of the first two terms: .$\displaystyle a + ar$
The fifth term is: .$\displaystyle a_5 \:=\:ar^4$

We have: .$\displaystyle a + ar \:=\:2ar^4 \quad\Rightarrow\quad 2r^4 - r - 1 \:=\:0$

. . which factors: .$\displaystyle (r-1)(2r^3 + 2r^2 + 2r + 1) \:=\:0$

We know that $\displaystyle r \neq 1$, and the cubic has no rational roots.

Now what?

4. Hello!
Thank you for helping me!

@ Soroban: I used the graphic calculator to solve that equation with power 4, and I could get 4 roots: 2 of which are the same (-0.17610), 1, and -0.64780...