Arithmetic Progression

$\underline{I.\ Introduction}$

A sequence of numbers is said to be in $Arithmetic\ Progression$ if the difference between any two consecutive terms is always constant.

For example, the series:

$3,\ 7,\ 11,\ 15...$ is an example of an arithmetic progression. The difference between any two consecutive term is $4$.

Likewise, the natural number series:

$1,\ 2,\ 3,\ 4...$ is also an arithmetic progression, and the difference between and two consecutive terms is $1$.

$\underline{II.\ Definitions\ \&\ Common\ Notations}$

The first term of an arithmetic progression is commonly denoted as $a$.

In the series $3,\ 7,\ 11,\ 15...$ $a = 3$

The constant difference between any term and its previous term is called the $common\ difference$ and is denoted by $d$.

In the series $3,\ 7,\ 11,\ 15...$ $d = 4$

The $n\xasuper{th}$ term is denoted by $T_n$ (read as $T\ sub\ n$ or $T n$)

In the series $3,\ 7,\ 11,\ 15...$ $T_3 = 11$

The sum of the first $n$ terms is denoted by $S_n$

Therefore, $S_n = T_1 + T_2 + ... + T_n$

In the series $3,\ 7,\ 11,\ 15...$ $S_3 = 3 + 7 + 11 = 21$

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$\underline{III.\ Terms\ Of\ An\ Arithmetic\ Progression}$

In this section we will see how to quickly find the $n\xasuper{th}$ term of a series.

$T_1 = a$

$T_2 = a + d$

$T_3 = T_2 + d = a + d + d = a + 2d$

$T_4 = T_3 + d = a + 2d + d = a + 3d$

$...$

$T_n = a + (n-1)d$

$\underline{IV.\ Sum\ Of\ Terms}$

Now we will see how to quickly find the sum of the first $n$ terms of an arithmetic progression.

$S_n = T_1 + T_2 + T_3 + ... + T_{n-2} + T_{n-1} + T_n$

$\Rightarrow S_n = a + (a + d) + (a + 2d) + ...$$\ \ \ + \{a + (n-3)d\} + \{a + (n-2)d\} + \{a + (n-1)d\}$ $...eqn\ (i)$

Since addition is commutative, we can write the right hand side of this equation in the reverse order, that is, the last term first and the first term last, and the sum would remain the same.

Therefore, we can write:

$S_n = T_n + T_{n-1} + T_{n-2} + ... + T_3 + T_2 + T_1$

$\Rightarrow S_n = \{a + (n-1)d\} + \{a + (n-2)d\} + \{a + (n-3)d\} + ...$$\ \ \ + (a + 2d) + (a + d) + a$ $...eqn\ (ii)$

Note, that the right hand side of both $eqn (i)$ and $eqn(ii)$ has the same number of terms.

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Now we can add $equations\ (i)$ and $(ii)$ and group the right hand side terms in pairs, that is one term from equation $(i)$ and one term from equation $(ii)$, we get:

$S_n + S_n = \underline{[a + \{a + (n-1)d\}]} + \underline{[(a + d) + \{a + (n-2)d\}]}$$\ \ \ + \underline{[(a + 2d) + \{a + (n-3)d\}]} + ... + \underline{[\{a + (n-1)d\} + a]}$

Each underlined group on the right hand side above, is a pair formed by taking one term from $eqn\ (i)$ and the corresponding term from $eqn\ (ii)$, and there will be $n$ such groups.

$\therefore 2S_n = \underline{[2a + (n-1)d]} + \underline{[2a + (n-1)d]} + \underline{[2a + (n-1)d]} + ... $$\ \ \ + \underline{[2a + (n-1)d]}$

Note that each group now has the same value, and we have $n$ such groups.

$\therefore 2S_n = n \times \{2a + (n-1)d\}$

$\Rightarrow S_n = \dfrac{n}{2} \{2a + (n-1)d\}$