Derivation of Sum of Arithmetic Progression

Arithmetic Progression, AP
Definition

Arithmetic Progression (also called arithmetic sequence), is a sequence of numbers such that the difference between any two consecutive terms is constant. Each term therefore in an arithmetic progression will increase or decrease at a constant value called the common difference, d.
 

Examples of arithmetic progression are:

  • 2, 5, 8, 11,... common difference = 3
  • 23, 19, 15, 11,... common difference = -4

 

Derivation of Formulas
Let
$d$ = common difference
$a_1$ = first term
$a_2$ = second term
$a_3$ = third term
...

$a_m$ = mth term or any term before $a_n$
...

$a_n$ = nth term or last term
 

$d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3$ and so on.
 

Derivation for an in terms of a1 and d
$a_1 = a_1$

$a_2 = a_1 + d$

$a_3 = a_2 + d = (a_1 + d) + d = a_1 + 2d$

$a_4 = a_3 + d = (a_1 + 2d) + d = a_1 + 3d$

$a_5 = a_4 + d = (a_1 + 3d) + d = a_1 + 4d$
...

$a_m = a_1 + (m - 1)d$
...

$a_n = a_1 + (n - 1)d$

 

In similar manner
$a_n = a_n$

$a_{n - 1} = a_n - d$

$a_{n - 2} = a_{n - 1} - d = (a_n - d) - d = a_n - 2d$

$a_{n - 3} = a_{n - 2} - d = (a_n - 2d) - d = a_n - 3d$

$a_{n - 4} = a_{n - 3} - d = (a_n - 3d) - d = a_n - 4d$
...

$a_m = a_n - (n - m)d$
...

$a_1 = a_n - (n - 1)d$

 

Derivation for the Sum of Arithmetic Progression, S
$S = a_1 + a_2 + a_3 + a_4 + ... + a_n$

$S = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + ... + [ \, a_1 + (n - 1)d \, ]$   →   Equation (1)
 

$S = a_n + a_{n - 1} + a_{n - 2} + a_{n - 3} + ... + a_1$

$S = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + ... + [ \, a_n - (n - 1)d \, ]$   →   Equation (2)
 

Add Equations (1) and (2)
$2S = (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + ... + (a_1 + a_n)$

$2S = n(a_1 + a_n)$

$S = \dfrac{n}{2}(a_1 + a_n)$

 

Substitute an = a1 + (n - 1)d to the above equation, we have
$S = \dfrac{n}{2} \{ \, a_1 + [ \, a_1 + (n - 1)d \, ] \, \}$

$S = \dfrac{n}{2}[ \, 2a_1 + (n - 1)d \, ]$

 

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