Derivation of Sum of Arithmetic Progression


A D V E R T I S E M E N T


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
[math]d[/math] = common difference
[math]a_1[/math] = first term
[math]a_2[/math] = second term
[math]a_3[/math] = third term
...
[math]a_m[/math] = mth term or any term before [math]a_n[/math]
...
[math]a_n[/math] = nth term or last term

[math]d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3[/math] and so on.

 

Derivation for [math]a_n[/math] in terms of [math]a_1[/math] and [math]d[/math]

[math]a_1 = a_1[/math]
[math]a_2 = a_1 + d[/math]
[math]a_3 = a_2 + d = (a_1 + d) + d = a_1 + 2d[/math]
[math]a_4 = a_3 + d = (a_1 + 2d) + d = a_1 + 3d[/math]
[math]a_5 = a_4 + d = (a_1 + 3d) + d = a_1 + 4d[/math]
...
[math]a_m = a_1 + (m - 1)d[/math]
...

a_n = a_1 + (n - 1)d

 

In similar manner
[math]a_n = a_n[/math]
[math]a_{n - 1} = a_n - d[/math]
[math]a_{n - 2} = a_{n - 1} - d = (a_n - d) - d = a_n - 2d[/math]
[math]a_{n - 3} = a_{n - 2} - d = (a_n - 2d) - d = a_n - 3d[/math]
[math]a_{n - 4} = a_{n - 3} - d = (a_n - 3d) - d = a_n - 4d[/math]
...
[math]a_m = a_n - (n - m)d[/math]
...

a_1 = a_n - (n - 1)d

 

Derivation for the Sum of Arithmetic Progression, S

[math]S = a_1 + a_2 + a_3 + a_4 + ... + a_n[/math]
[math]S = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + ... + [ \, a_1 + (n - 1)d \, ] \, \to \, [/math] Equation (1)

[math]S = a_n + a_{n - 1} + a_{n - 2} + a_{n - 3} + ... + a_1[/math]
[math]S = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + ... + [ \, a_n - (n - 1)d \, ] \, \to \, [/math] Equation (2)

 

Add Equations (1) and (2)
[math]2S = (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + ... + (a_1 + a_n)[/math]
[math]2S = n(a_1 + a_n)[/math]

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

 

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

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



Page copy protected against web site content infringement by Copyscape