# Solve Geometric Series: Find n from s=9-32-n

• MHB
• ChelseaL
In summary, we have proven that the sum of the first n terms of series s, 9-32-n, is a geometric progression with a common ratio of 1/3. This was done by using the formula an = ar n-1 and showing that the ratio of consecutive terms is always 1/3. Thus, the sum s is a geometric progression.
ChelseaL
Given that the sum of the first n terms of series, s, is 9-32-n
show that the s is a geometric progression.

Do I use the formula an = ar n-1? And if so, how do I apply it?

Last edited:
So far I have this

9-32-n

3 = 9
= 9 -(n-2)
= 9 -(n-1-1)
= 9 -(n-1) + 1
= 9 \cdot 3 -(n-1)
= 9 (\frac{1}{3}) n-1
where a=9 and r=\frac{1}{3}

We have:

$$\displaystyle S_{n}=S_{n-1}+a_{n}\implies a_{n}=S_{n}-S_{n-1}=\left(9-3^{2-n}\right)-\left(9-3^{2-(n-1)}\right)=\frac{18}{3^n}$$

To show that $a_n$ is a GP, we need to show that the ratio $$\displaystyle \frac{a_{n+1}}{a_{n}}$$ is a constant, that is, not dependent on $n$.

Can you proceed?

What do you mean?

ChelseaL said:
What do you mean?

show the ratio $\dfrac{a_{n+1}}{a_n}$ is a constant

18/3n + 1 / (18/3n)?

Sorry for not really understanding this topic. My lecturer regularly misses class and refuses to help us when we ask questions.

ChelseaL said:
18/3n + 1 / (18/3n)?

Sorry for not really understanding this topic. My lecturer regularly misses class and refuses to help us when we ask questions.

If $$\displaystyle a_n=\frac{18}{3^n}$$ then $$\displaystyle a_{n+1}=\frac{18}{3^{n+1}}$$ and so:

$$\displaystyle \frac{a_{n+1}}{a_{n}}=\frac{\dfrac{18}{3^{n+1}}}{\dfrac{18}{3^{n}}}=\,?$$

\frac{1}{n+1}?

ChelseaL said:
\frac{1}{n+1}?

No, I think you are not using proper rules of algebra. Observe that:

$$\displaystyle a_{n+1}=\frac{18}{3^{n+1}}=\frac{18}{3\cdot3^n}=\frac{1}{3}a_n$$

Hence:

$$\displaystyle \frac{a_{n+1}}{a_{n}}=\frac{1}{3}$$

ohhhh. What do I need to do from here?

ChelseaL said:
ohhhh. What do I need to do from here?

As far as this problem goes, it is done, because we have shown the ratio is a constant. :)

## What is a geometric series?

A geometric series is a sequence of numbers where each term is found by multiplying the previous term by a constant value. For example, 2, 6, 18, 54, ... is a geometric series where the constant value is 3.

## How do I solve a geometric series?

To solve a geometric series, you can use the formula: S = a(1-r^n)/(1-r), where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms. Rearrange the formula to solve for n by isolating it on one side.

## What is the common ratio in a geometric series?

The common ratio in a geometric series is the constant value that is multiplied to each term to get the next term. It is usually denoted as 'r' and can be found by dividing any term by its previous term.

## Is it possible to have a negative number of terms in a geometric series?

No, it is not possible to have a negative number of terms in a geometric series. The number of terms, denoted as 'n', represents the position of the last term in the series. It is always a positive integer.

## Can I use a calculator to solve a geometric series?

Yes, you can use a calculator to solve a geometric series. Most scientific calculators have a function for finding the sum of a geometric series. You can also use online calculators or spreadsheet programs like Excel.

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