Apakah itu barisan geometrik?

(Video) Geometric Progressions

by Eddie Woo

Here's some videos by Eddie Woo going over geometric series and sequences.

Recall that each successive term in an arithmetic progression has a common difference, $d$.

In this video he goes over: $$\\\begin{align} &\circ\text{ Geometric Progressions}\\ &\circ\text{ Sum of a Geometric Series}\\ &\circ\text{ Proving a Geometric Series} \end{align}$$

Geometric Progressions

A geometric progression takes the idea of an algebraic progression and modifies it slightly. The idea for a geometric progression is to have a common multiplication or a common division. In other words, each successive term in a geometric progression has a common ratio, $r$.

For example, when we listed all the even numbers, we saw they got bigger by two every time. Now what if we multiply by two instead? This would be a geometric progression. If we were to start at 1, the series would look like: $$\begin{array}{rrrrrc} 1+&2+&4+&8+&16+&\cdot\cdot\cdot\\ (1\times2^0)+&(1\times2^1)+&(1\times2^2)+&(1\times2^3)+&(1\times2^4)+&\cdot\cdot\cdot \end{array}$$

When we wrote a general series for arithmetic progression we got: $$a_1+(a_1+d)+(a_1+2d)+\cdot\cdot\cdot$$ A general series for geometric progression is similar. $$a_1+a_1r+a_1r^2+a_1r^3+\cdot\cdot\cdot$$

The arithmetic progression $n^{\text{th}}$ term looked like: $$a_n=a_1+(n-1)d$$ For the geometric progression, the $n^{\text{th}}$ term looks like: $$a_n=a_1r^{n-1}$$

(Video) Geometric Progressions

by Eddie Woo

Sum of a Geometric Series

With arithmetic progressions, the sum of the $n^{\text{th}}$ term is: $$S_n=\frac{n}2[2a_1+(n-1)d]$$

Now what about the geometric progression? We know the series looks like: $$S_n=a_1+a_1r+a_1r^2+\cdot\cdot\cdot+a_1r^{n-3}+a_1r^{n-2}+a_1r^{n-1}$$ To turn this into a more digestible form, we're going to multiply through by the common ratio, $r$. $$rS_n=a_1r+a_1r^2+a_1r^3+\cdot\cdot\cdot+a_1r^{n-2}+a_1r^{n-1}+a_1r^n$$

If we look at these above one another we see that they have some common terms. $$S_n=a_1+a_1r+a_1r^2+\cdot\cdot\cdot+a_1r^{n-3}+\boxed{a_1r^{n-2}}+\boxed{a_1r^{n-1}}\\ rS_n=a_1r+a_1r^2+a_1r^3+\cdot\cdot\cdot+\boxed{a_1r^{n-2}}+\boxed{a_1r^{n-1}}+a_1r^n$$ Actually all the terms are common except for two. $$\begin{alignat}{2} S_n&=\color{#8925AE}{\boxed{a_1}}+&&a_1r+a_1r^2+\cdot\cdot\cdot+a_1r^{n-3}+a_1r^{n-2}+a_1r^{n-1}\\ rS_n&=&&a_1r+a_1r^2+a_1r^3+\cdot\cdot\cdot+a_1r^{n-2}+a_1r^{n-1}+\color{#8925AE}{\boxed{a_1r^n}} \end{alignat}$$ Now we want to subtract these two. When we do this all the common terms will end up canceling. $$\begin{align} rS_n-S_n&=a_1r^n-a_1\\ S_n(r-1)&=a_1(r^n-1)\\ S_n&=\frac{a_1(r^n-1)}{r-1} \end{align}$$ This is the way we write the sum formula for the $n^{\text{th}}$ term.

Some people or textbooks write it differently but they mean the same thing. $$S_n=\frac{a_1(r^n-1)}{r-1}\quad\Leftrightarrow\quad S_n=\frac{a_1(1-r^n)}{1-r}$$

Suppose now we had a geometric series that was getting smaller. Let's think about how we'd find the sum. $$1+\frac12+\frac14+\frac18+\cdot\cdot\cdot$$ This series is known as a limiting sum or finite sum. It reaches a number and doesn't continue any further, where as when we multiplied by $2$ that went up to infinity (an infinite sum).

Proving a Geometric Progression

To prove an arithmetic progression we wanted to show that $$a_{n+1}-a_n=a_n-a_{n-1}$$

Instead of subtraction, to prove that something is a geometric progression we want to use division. $$\frac{a_{n+1}}{a_n}=\frac{a_n}{a_{n-1}}$$

(Video) A Quick Intro to Geometric Sequences

by PatrickJMT

This video by PatrickJMT teaches about geometric sequences and the common ratio.

$\large{\text{Summary}}$

A geometric sequence is a sequence of numbers where each term after the first term is found by multiplying the previous one by a fixed non-zero number, called the common ratio.

Usually we let $r=\text{common ratio}$.

Example 1

$$3,6,12,24,48,96,\cdots$$

Notice that we multiply by $2$ to get from $3$ to $6$. Then, we multiply by $2$ again to get from $6$ to $12$. You have to multiply by $2$ to get to each next term. Therefore, this is a geometric sequence and the common ratio is $r=2$.

Example 2

$$3,\frac32,\frac34,\frac38,\frac3{16},\frac3{32},\frac3{64},\cdots$$

Notice that we multiply by $\frac12$ to get from $3$ to $\frac32$. Then, we multiply by $\frac12$ again to get from $\frac32$ to $\frac34$. You have to multiply by $\frac12$ to get to each next term. Therefore, this is a geometric sequence with a common ratio of $r=\frac12$.

Example 3

$$10,15,20,25,30,\cdots$$

This sequence isn't looking geometric. In fact, it seems arithmetic since each term increases by $5$. One way to check is to look at the ratios between two sets of consecutive terms. $$\frac{15}{10}=\frac32\\ \frac{20}{15}=\frac43$$ They don't have the same ratio. Therefore, this sequence isn't geometric.

Example 4

$$-1,0.1,-0.01,0.001,-0.0001$$

This sequence is hard to tell. If you ever have trouble, just check the ratios. $$\frac{0.1}{-1}=-\frac{1}{10}\\ \frac{-0.01}{0.1}=-\frac{1}{10}$$ They have the same ratio! Make to check the other term as well. But for speed, I'm going to spoil that this sequence is geometric! The common ratio is $r=-\frac1{10}$.

Maximum Depth Exceeded: Fun with Sequences and Python, Part 4

To read part 3 of this story, go here.

A geometric sequence is one for which you multiply the same number to any term in the sequence to get the next one. Let's build a geometric sequence $b_0, b_1, b_2,...$ by picking a value of $b_0$ and then building a recurrence formula. My lucky number is $11,$ so let's set $b_0 = 11.$

Now we can pick a number to multiply by. Let's try $3.$ This means for any term in the sequence, we multiply it by $3$ to get the next one, so its recurrence formula is

$$b_n = 3 \times b_{n - 1}.$$

We take the value of $b_0$ and the recurrence formula and build it into a Python function:

If we plug in $b(n)$ for different values of $n,$ we'll generate the sequence.

To see what happens, go to the next explanation in the series here.

Geometric Sequences

With geometric sequences, you multiply each previous term by a common ratio. For example, $$2, 4, 8, 16, 32...$$ , where the common ratio is: $2$.

Here is the formula for geometric sequences:

The geometric sequence formula is: $$a_n=a_1r^{n-1}$$

Let's look at an example.

Solution:

From the problem, we know that: $$n=4$$ $$a_1=2$$ $$r=5$$ The formula for geometric sequences is: $$a_n=a_1r^{n-1}$$ Let's plug in what we know. $$a_n=(2)5^{4-1}$$ $$a_n=(2)5^3$$ $$a_n=(2)(125)$$ $$a_n=250$$ Therefore, the $4th$ term is $250$.