# Arithmetic progression

Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum.

This sum can be found quickly by taking the number *n* of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2:

To derive the above formula, begin by expressing the arithmetic series in two different ways:

Adding the corresponding terms of both sides of the two equations and halving both sides:

The formula is very similar to the mean of a discrete uniform distribution.

The standard deviation of any arithmetic progression can be calculated as