# 22. Arithmetic Progressions¶

Consider sequences like 1, 2, 3, 4, … or -1, -2, -3, -4, … or 1, 3, 5, 7, … or $$a, a + d, a + 2d, a + 3d, ...$$

These sequences increase or decrease with a common difference. When quantities increase or decrease with a common difference they are said to be in Arithmetic Progression. The common difference can be found by subtracting any term of the series that follows it. For example for the first series it is 1 and for the last it is $$d$$.

Consider the series $$a, a + d, a + 2d, a + 3d, ...$$

Simple observation tells us the 1st term is $$a,$$ the 2nd term is $$a + d,$$ the third term is $$a + 2d$$ and hence the nth term will be $$a + (n - 1)d.$$ These terms are typically written as $$t_1, t_2, t_3, ..., t_n.$$

## 22.1. nth Term of Arithmetic Progression¶

Following above discussion we can clearly say that nth term of an arithmetic progression is given by $$t_n = a + (n - 1)d$$ where $$a$$ is called the first term and $$d$$ the common difference.

## 22.2. Sum of An Arithmetic Progression¶

Let $$S_n$$ represents the sum of first $$n$$ terms of an arithmetic progression. Then we can write

$S_n = a + (a + d) + (a + 2d) + ... + (a + (n - 2)d) + (a + (n - 1)d)$

Writing the terms in reverse order we have

$S_n = (a + (n - 1)d) + (a + (n - 2)d) + ... + (a + d) + a$

\begin{align}\begin{aligned}2S_n = [2a + (n - 1)d] + [2a + (n - 1)d] + ... \text{ to n terms}\\2S_n = n[2a + (n - 1)d]\\S_n = \frac{n}{2}[2a + (n - 1)d]\\S_n = \frac{n}{2}(a + t_n)\end{aligned}\end{align}

## 22.3. Arithmetic Mean¶

When three quantities are in arithmetic progression the quantity in the middle is known to be arithmetic mean of the other two. For example, if $$a, b, c$$ are in A. P. then $$b$$ is said to be arithmetic mean of $$a$$ and $$c.$$ In general it is written as $$b = \frac{a + c}{2}.$$ This can be examined further. Let $$b = a + d$$ then $$c = a + 2d.$$ Clearly, $$b = \frac{a + c}{2}.$$

It is also possible to insert $$n$$ numbers between any two numbers such that all of them are in A. P. Consider two number $$a$$ and $$b$$ in between which we want to insert $$n$$ numbers such that they are in A. P. Clearly,:math:b will become n + 2th term of A. P. Let common difference be $$d$$ then we can write $$b = a + (n + 1)d$$ or $$d = \frac{b - a}{n + 1}.$$ Now all the $$n$$ arithmetic means can be deduced. Let those are $$m_1, m_2, ..., m_n$$ then $$m_1 = a + \frac{b - a}{n + 1}, m_2 = a + \frac{2(b - a)}{n + 1}, ..., m_n = a + \frac{n(b - a)}{n + 1}.$$

Suppose there are $$n$$ terms of an A. P. then the arithmetic mean is given by

$\frac{t_1 + t_2 + ... + t_n}{n}.$

### 22.3.1. $$n$$ arithmetic means between two quantities¶

Let the two quantities be $$a$$ and $$b$$. Let $$x_1, x_2, ..., x_n$$ be those $$n$$ means. Then, $$b$$ becomes $$(n+2)nd$$ term. Thus,

$b = t_{n + 2} = a + (n + 1)d$
$d = \frac{b - a}{n + 1}$
$\therefore x_1 = a + d, x_2 = a + 2d, ..., x_n = a + nd$

Substituting for $$d$$

$x_1 = \frac{na + b}{n + 1}, ..., x_n = \frac{nb + a}{n + 1}$

## 22.4. Deducing Number of Terms¶

Now we know that $$S_n = \frac{n}{2}[2a + (n - 1)d]$$ say $$S_n, a$$ and $$d$$ are known and we have to evaluate $$n.$$ This being a quadratic equation in $$n$$ we will have two roots for $$n.$$ If the results are positive and integral then there is no difficulty in interpreting the results for them. In some cases for a negative root a suitable interpretation can be given.

Example How many terms of the series -8, -6, -4, … must be added that the sum be 36?

Here $$\frac{n}{2}[-16 + (n - 1)2] = 36 \Rightarrow 2n^2 - 18n - 72 = 0 \Rightarrow n^2 - 9n - 36 = 0 \Rightarrow n = 12 or -3.$$

If we take 12 terms of the series, we have -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14 the sum of which is 36. The sum of last 3 elements is also 36. Thus the negative solution even though does not directly answer the question, we can give it a meaning.

## 22.5. Some Observations on an A. P.¶

1. If a fixed number is added to or subtracted from each term of a given A. P., then the resulting sequence is also an A. P., and it has the same common difference as that of the given A. P.

2. If each term of an A. P. is multiplied or divided by a non-zero fixed constant then the resulting sequence is also an A. P.

3. If $$a_1, a_2, a_3, ...$$ and $$b_1, b_2, b_3, ...$$ are two arithmetic progressions then $$a_1 + b_1, a_2 + b_2, a_3 + b_3, ...$$ are also in A. P.

4. If we have to choose three unknown terms in an A. P. then it is best to choose them as $$a -d, a, a + d$$.

5. If we have to choose four unknown terms in an A. P. then it is best to choose them as $$a - 3d, a - d, a + d, a + 3d$$.

6. In an A. P. the sum of terms equidistant from the beginning and end is constant and equal to the sum of first and last term.

7. Any term of an A. P., except the first, is equal to half the sum of terms which are equidistant from it:

$a_n = \frac{1}{2}(a_{n - k} + a_{n + k}),~~~~~k<n, \text{and for } k = 1,$
$a_n = \frac{1}{2}(a_{n - 1} + a_{n + 1})$
8. $$t_n = S_n - S_{n - 1}, ~~(n \ge 2)$$

9. If $$t_n = pn + q$$ i.e. a linear expression in $$n$$ then it will form an A. P. of common difference $$p = t_n - t_{n - 1}$$ and first term $$p + q~(n =1)$$ i.e. if $$t_n = 3n + 4$$, then it is an A. P. of common difference 3 and first term as 7.

10. If $$S_n = an^2 + bn + c$$ i.e. quadratic function in $$n$$ then the series is A. P. where $$d = 2a$$, twice the coefficient of $$n^2$$.

## 22.6. Sum of Squares and Cubes and More¶

We observe that

\begin{align}\begin{aligned}i^3 - (i - 1)^3 = 3i^2 - 3i + 1\\\sum_{i=1}^n i^3 - (i - 1)^3 = 3\sum_{i=1}^n i^2 - \frac{3n(n + 1)}{2} + n\\n^3 = 3\sum_{i=1}^n i^2 - \frac{3n(n + 1)}{2} + n\\3\sum_{i=1}^n i^2 = n^3 + \frac{3n(n + 1)}{2} - n\\\sum_{i=1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}\end{aligned}\end{align}

Following in a similar fashion we can show that

$\sum_{i=1}^n i^3 = \left\{\frac{n(n + 1)}{2}\right\}^2$

More powers can be evaluated in a similar fashion.