User:Extra999/AP

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Arithmetic progressions
The difference between any two terms of the series is a constant, called common difference.

For example,

2,5,8,11,14,...

1,2,3,4,...

-10,-5,0,5,10,...

If the first term is denoted by a and common difference by d.

then series is given by:

a,a+d,a+2d,...

Therefore the nthis given by a+(n-1)d

Finding the sum of an arithmetic progression
Let the sum be denoted by S

$$S=(a)+(a+d)+(a+2d)+(a+3d)+(a+4d)...+a+(n-1)d$$

also $$S=(a+(n-1)d)...+(a+d)+a$$

Adding these we get

$$2S=(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)...$$ n times

Therefore $$S=n(2a+(n-1)d)/2$$