Arithmetic Progression

In the previous post, we talked a little about the history and ideas about arithmetic progression. In this post, we will delve deeper into the concept.

Arithmetic progression

Definition:

An arithmetic progression (A.P.) is called a sequence given by the following recurrence formula:

where $a$ and $r$ are given real numbers.

Thus, an A.P. is a sequence in which each term, starting from the second, is the sum of the previous one with a given constant $r$.

Here are some examples of arithmetic progressions:

Classification

Arithmetic progressions can be classified into three categories:

1. Crescent are the A.P. in which each term is greater than the previous one. AND It is immediate that this only occurs if $r > 0$, because:

Examples: $f_1$ and $f_4$.

2. Constants are the A.P. in which each term is equal to the previous one. It’s easy to see This only occurs when $r = 0$, because:

Example: $f_3$.

3. Decreasing are the A.P. in which each term is smaller than the previous one. This only occurs if $r < 0$, because:

Examples: $f_2$ and $f_5$.

Special notations

When we seek to obtain a P.A. with 3, 4 or 5 terms, it is very practical the following notation:

1. for 3 terms: $(x, x + r, x + 2r)$ or $(x - r, x, x + r)$
2. for 4 terms: $(x, x + r, x + 2r, x + 3r)$ or $(x - 3y, x - y, x + y, x + 3y)$, where $y = \frac{r}{2}$
3. for 5 terms: $(x, x + r, x + 2r, x + 3r, x + 4r)$ or $(x - 2r, x - r, x, x + r, x + 2r)$

Exercise: Determine $x$ so that $(x, 2x + 1, 5x + 7)$ is a P.A.

Solution

We should have $a_2 - a_1 = a_3 - a_2$. Then:

Exercise: Determine $a$ so that $(a^2, (a + 1)^2, (a + 5)^2)$ is a P.A.

Solution

We must have $a_2 - a_1 = a_3 - a_2$, that is, $2a_2 = a_1 + a_3$. Then:

Exercise: Obtain an A.P. of three terms such that their sum is 24 and their product is 440.

Solution

Using the special notation $(x - r, x, x + r)$ for P.A., we have:

From (*) we obtain $3x = 24$, that is, $x = 8$. Replacing in (**), we get:

Thus, the P.A. sought is:

$(5, 8, 11)$ to $x = 8$ and $r = 3$, or $(11, 8, 5)$ to $x = 8$ and $r = -3$.

General term formula

Using the recurrence formula by which a P.A. is defined and assuming given the first term ($a_1$), the ratio ($r$) and the index ($n$) of a desired term, we have:

Adding these $n-1$ equalities, we have:

and then $a_n = a_1 + (n-1) ~.~ r$, which suggests the following

Theorem

In A.P. where the first term is $a_1$ and the ratio is $r$, the nth term is

Proof by the principle of finite induction

I) For $n = 1$, we have: $a_1 = a_1 + (1-1) ~.~ r$ (true sentence).

II) Let us admit the validity of the formula for $n = p$:

and we prove that it holds for $n = p + 1$:

So, $a_n = a_1 + (n-1) ~.~ r, \ \forall n \in \mathbb{N}^*$.

Exercises

1. Calculate the 17th term of the A.P. whose first term is 3 and whose ratio is 5.

Solution

Noting that $a_1 = 3$ and $r = 5$, let us apply the general term formula:

2. Obtain the ratio of A.P. where the first term is $-8$ and the twentieth is $30$.

Solution

3. Get the P.A. at which $a_{10} = 7$ and $a_{12} = -8$.

Solution

To write the P.A. it is necessary to determine $a_1$ and $r$. We have:

Solving the above system, we have:

1. $(**) - (*)$: $\ 2r = -15 \quad\Rightarrow\quad r = -\frac{15}{2}$
2. in $(*)$: $\ a_1 + 9 ~.~ \left(-\frac{15}{2}\right) = 7 \quad\Rightarrow\quad a_1 = \frac{149}{2}$

and therefore the P.A. is $\left(\frac{149}{2}, \frac{134}{2}, \frac{119}{2}, \dots\right)$.

Arithmetic Interpolation

In every finite sequence $(a_1, a_2, \dots, a_{n-1}, a_n)$, the terms $a_1$ and $a_n$ are called extremes and the others are called middles. Thus, in P.A. $(0, 3, 6, 9, > 12, 15)$ the extremes are $0$ and $15$, while the means are $3, 6, 9$ and $12$.

Interpolate, insert or interleave $k$ arithmetic means between the numbers $a$ and $b$ means obtaining a P.A. of extremes $a_1 = a$ and $a_n = b$, with $n = k + 2$ terms. To determine the means of this P.A. it is necessary to calculate the ratio, which is done like this:

Example: Interpolate 5 arithmetic means between 1 and 2.

Let’s form a P.A. with 7 terms where $a_1 = 1$ and $a_7 = 2$. We have:

so the P.A. is

Sum

Let’s derive a formula to calculate the sum $S_n$ of the $n$ initial terms of a P.A.

Theorem 1

The sum of $n$ first positive integers is $\frac{n(n+1)}{2}$.

Proof by finite induction

I) For $n = 1$, we have: $1 = \frac{1(1 + 1)}{2}$ (true sentence).

II) Let us admit the validity of the formula for $n = p$:

and we prove that it holds for $n = p + 1$:

Then $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}, \ \forall n \in \mathbb{N}^*$.

Example: The sum of the 50 initial terms of the sequence of positive integers is:

Using the general term formula, we can calculate the sum $S_n$ of the $n$ initial terms of the P.A. $(a_1, a_2, \dots, a_n, \dots)$.

Theorem 2

In all P.A. there is: $S_n = na_1 + \frac{n(n-1)}{2} ~.~ r$

Demo

Adding these equalities:

By the Theorem 1, $1 + 2 + \dots + (n-1) = \frac{(n-1)n}{2}$; then,

that is:

Theorem 3

In all P.A. there is: $S_n = \frac{n(a_1 + a_n)}{2}$

Demo

Examples

1. The sum of the 15 initial terms of A.P. $(-2, 1, 4, 7, \dots)$ is:

2. The sum of the integer multiples of 2 from 4 to 100 can be calculated by noting that $(4, 6, 8, \dots, 100)$ is an A.P. of 49 terms where $a_1 = 4$ and $a_{49} = 100$:

Laboratory: recognize, calculate and check

In the training below, three missions appear: discover the reason, calculate a term and add terms. Write down your strategy before checking. The difficulty changes the number of terms considered. The questions in this activity use natural terms and ratios, including zero.

Challenge: when do terms stop being natural?

A PA starts at 9 and has reason −2. Write the first six terms. Up to what position are all terms still natural?