# Sum of First n Terms of an AP

#### 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)
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 
S100= 11300
∴ Sum of first 100 terms of given A.P. is 11,300

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