#### description

- Sum of the First 'N' Terms of an Arithmetic Progression

#### formula

Sum of first n terms of an `"AP": "S" =(n/2)[2a + (n- 1)d]` The sum of n terms is also equal to the formula where l is the last term.

#### notes

Arithmetic Progression a, a + d, a + 2d, a + 3d, . . . . . . . . . . . . a +(n - 1)d

In this progression a is the first term and d is the common difference. Let’s write the sum of first n terms as Sn.

Sn = [a] + [a + d] + . . . + [a+(n-2)d] + [a+(n-1)d] ..........(eq1)

Reversing the terms and rewritting the expression again,

Sn = [a+(n-1)d] + [a+(n-2)d] + . . . + [a + d ] + [a] ........(eq2)

On adding eq1 and eq2

2Sn = [a+a+(n-1)d] + [a + d+a+(n-2)d]+ . . . + [a+(n-2)d+ a + d]+ [a+(n-1)d+a]

2Sn = [2a+(n-1)d] + [2a+(n-1)d] + . . . + [2a+(n-1)d] . . . n times.

∴2Sn = n [2a+(n-1)d]

∴Sn = n/2 [2a+(n-1)d]

Ex. Let’s find the sum of first 100 terms of A.P. 14, 16, 18, . . . .

Here a= 14, d = 2, n = 100

`Sn = n/2 [2a+(n-1)d]`

`S100= 100/2 [2× 14+ (100-1)2]`

`S100= 50 [28 + (99)2]`

`S100= 50 [28 + 198]`

`S100= 50 [226]`

`S100= 11300`

∴ Sum of first 100 terms of given A.P. is 11,300