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 n^this 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

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