# Sum of n terms of an Arithmetic Progression

Refer to an actual incident explained at the right.
We will now use the same technique to find the sum of the first "n" terms of an A.P.
Let the first term of an A.P. be a and common difference be d.
Let Sn denote the sum of the n terms of A.P. Then
Sn = a + [a + d] + [a + 2d] + . . . . + [a + (n – 2)d] + [a + (n – 1)d]
Rewriting the terms in reverse order, we have
Sn = [a + (n – 1)d] + [a + (n – 2)d] + . . . . + [a + 2d] + [a + d] + a
On adding the two
Sn + Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + . . . . . . .+ [2a + (n – 1)d] +
[2a + (n – 1)d] (n times)
⇒ 2Sn = n[2a + (n – 1)d]
Sn = [2a + (n – 1)d]
We can also write
Sn = [a + a + (n – 1)d] = [a + an]   since a + (n – 1)d = an
Sometimes Sn is simply denoted by S.
If there are only 'n' terms in an A.P. and an = l (the last term), then
S = [a + l]
The above formula is useful when the first and last terms of the A.P. are known.

In solving problems related to Arithmetic progressions, the following shall be useful:
i. 3 successive terms of an A.P. can be considered as: (a – d), a, (a + d)
ii. 4 successive terms of an A.P. can be considered as: (a – 3d), (a – d), (a + d), (a + 3d)
iii. 5 successive terms of an A.P. can be taken as: (a – 2d), (a – d), a, (a + d), (a + 2d)
iv. The nth term of an A.P. is equal to sum of n terms minus sum of (n – 1) terms
i.e, Tn = Sn – Sn-1
v. The sum of terms equidistant from beginning and end is constant