Sequences and series

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

Geometric progressions
Ratio of any two terms of the series is constant. If first term is denoted by a, and common ratio by r. Then the series is given by:- a,ar,ar2,...ar(n-1) Example:- 1,2,4,8,16,... 1,-2,4,-8,16,...(note here that r is negative)

Finding the sum of a geometric progression
Let the sum be denoted by S S=a+ar+... (i) Multiply the equation by r. rS=ar+ar2... (ii) Subtract (ii) from (i) S(1-r)=a-arn This gives S=a(1-rn)/(1-r) (for r not equal to 1)

When r=1, S=a+a+a+...(to n terms) S=na