等比数列

Let's see if we can find the formula for finding the nth term in a geometric sequence. Let's take a look at a pretty basic sequence first: $$2,4,8,16,\dots$$ It's pretty obvious that the ratio between terms is simply $2$. Furthermore, you might notice that each of the terms is simply a power of $2$. The first term is $2^1$, the second term is $2^2$, and so on. It seems pretty clear that the nth term is $2^n$.

Now let's look at another sequence, $$5,15,45,135,\dots$$ We should immediately notice that the common ratio is $3$. We can write all of the terms in an alternate form, $$5\cdot 3^0,5\cdot 3^1,5\cdot 3^2,\dots$$ This makes it pretty obvious that this sequence can be written as $5\cdot 3^{n-1}$. It is a pretty common mistake to accidentally put $5\cdot 3^{n}$ as the formula for the nth term (instead of $5\cdot 3^{n-1}$) so make sure not to fall into that trap! Remember that $5$ is the first term, not the zeroth term.

Let's take another look at that first sequence. We can also write this sequence in a similar form: $$2\cdot 2^0,2\cdot 2^1,2\cdot 2^2,\dots$$ It looks like the nth term is just $2\cdot 2^{n-1}$. But we got $2^n$ above! Remember that $2\cdot 2^{n-1}=2^n$. So we have come up with the same formula, just in a different form.

After looking at these examples, you may have come up with an idea for a formula. First, let's define some variables.

Let $a$ be the first term in your sequence, and let $r$ be the ratio between consecutive terms. Looking at our examples above it seems like the formula for the nth term is just $a\cdot r^{n-1}$.

Let's try an example.

Geometric Sequences

Intuitively, a geometric sequence is a sequence where a term in the sequence is found by multiplying the previous term by some constant number.

More formally, a geometric sequence is a sequence of the form:

$$a,\;ar,\; ar^2,\; ar^3,\;\ldots$$

where $a$ is the first term in the sequence, and $r$ is the common ratio. (Note that the common ratio is the same throughout the sequence.)

Finding the general term

Often it's useful to find the general term of a geometric sequence. The general term is the algebraic formula for the value of the $n^{\text{th}}$ term in the sequence.

We can see the general term by looking at a general geometric sequence. If we let the first term be $a$, and the common ratio be $r$, then: $$a,\;ar,\; ar^2,\; ar^3,\;\ldots$$ It follows from this that the $n^{th}$ term is given by $ar^{n-1}$. (Follow the value of the exponents carefully, and you will see that power of $r$ in the first term is $0$, in the second term it's $1$, in the third it's $2$, etc.). We often write this as: $$T_n = ar^{n-1}$$ This could be thought of as meaning "the formula for term number $n$ is $ar^{n-1}$".

Finding the value of a term given the general term

Once we have an expression for the general term, it's easy to calculate the value of any particular term in the sequence.

For example, if we are given that $T_n = 4\cdot 3^{n-1}$, we can calculate the value of the tenth term by replacing $n$ in the formula with $10$: $$\begin{align} T_{10} & = 4\cdot 3^{10-1} \\ & = 4\cdot 3^{11} \\ & = 708\, 588 \end{align}$$ Exercises

Finding the number of a term given a value

If we have the general term, we can also find out which term number in the sequence corresponds to a certain value.

For example, if you are asked which term in the sequence given by $T_n=3\cdot (\frac{1}{2})^{n-1}$ has a value of $\frac{3}{16}$.

We solve for $n$ in the following equation:

$$\begin{align} 3\cdot \left(\frac{1}{2}\right)^{n-1} & = \frac{3}{16} \\ \left(\frac{1}{2}\right)^{n-1} & = \frac{1}{16} \\ \frac{1}{2^{n-1}} & = \frac{1}{2^4} \\ \therefore n & = 5 \end{align}$$

Thus we can see that the fifth term in the sequence has a value of $\frac{3}{16}$.

(Video) Hard Arithmetic and Geometric Progression Questions

by Eddie Woo

Here's a video by Eddie Woo where he goes over some harder questions involving arithmetic progressions and geometric progressions.

If you need a quick reminder of sequences, click here.

The first question he looks at asks:

Show that if the first, second, and fourth terms of an arithmetic progression form a geometric sequence. If that's the case, show that either (1) the sequence is a constant sequence or (2) that the terms are the positive integer multiples of the first term.

We start by writing the general form of an arithmetic progression. There's the first term $a$ and then each term after differs by a difference of $d$. $$a,a+d,a+2d,a+3d,\dots$$ The question is interested in the first, second, and fourth terms. $$\boxed{\color{red}{a}},\boxed{\color{red}{a+d}},a+2d,\boxed{\color{red}{a+3d}},\dots$$ It is saying that these three form a geometric sequence. To prove that something is a geometric progression you need to show that: $$\frac{T_{n+1}}{T_{n}}=\frac{T_n}{T_{n-1}}$$ So we want to prove that: $$\frac{a+d}{a}=\frac{a+3d}{a+d}$$

Before we continue proving the above let's understand what the two cases at the end mean.

Constant Sequence

A constant sequence is a when the sequence as a whole is constant. That means all the terms are the same. $$a,a,a,a,a,\dots$$ This would mean, for the arithmetic progression, the difference would be zero, $d=0$.

Positive Integer Multiples

If the terms are the positive integer multiples of the first then that means they're the positive integer multiples of $a$. $$a,2a,3a,4a,\dots$$ We can rewrite this as how we define an arithmetic sequence. $$a,a+a,a+2a,a+3a,\dots$$ In this case, the common difference is the first term: $d=a$.

So what we want to do is prove that either:

1. $d=0$ or
2. $d=a$

So now we go back to our equation and attempt to solve for $d$. $$\begin{align} \frac{a+d}{a}&=\frac{a+3d}{a+d}\\ (a+d)(a+d)&=a(a+3d)\\ a^2+2ad+d^2&=a^2+3ad\\ d^2-ad&=0\\ d(d-a)&=0 \end{align}$$

Try this!

The first, third, and fourth terms of an increasing geometric sequence form an arithmetic progression. What is the common ratio of the sequence?

Let the first term be $a$ and the common ratio be $r$. Then the first, third, and fourth terms are $a$, $ar^2$, and $ar^3$. Since they form an arithmetic sequence, we know $$ar^3-ar^2=ar^2-a.$$ We can factor both sides to get $$a(r-1)r^2=a(r-1)(r+1).$$ Since the arithmetic progression is increasing, we know that $a \neq 0$ and $r \neq 1$, so we can cancel $a(r-1)$ from both sides to get $$r^2=r+1.$$ Now solving the quadratic yields $r = \dfrac{1\pm\sqrt{5}}{2}$, but since the sequence is increasing, $r$ must be positive, and thus we get $r = \boxed{\dfrac{1+\sqrt{5}}{2}}$.

A geometric series or geometric progression is a series in which the ratio of any two consecutive terms is the same. This ratio is called the common ratio of the geometric progression, labeled $r$.

For example, if $a,b,c,d,\dots$ are terms of a geometric progression, then $$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}.$$