Sequences and series

## Arithmetic progressions

[edit | edit source]

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

[edit | edit source]

Let the sum be denoted by S
S = ( a ) + ( a + d ) + ( a + 2 d ) + ( a + 3 d ) + ( a + 4 d ) . . . + a + ( n − 1 ) d {\displaystyle S=(a)+(a+d)+(a+2d)+(a+3d)+(a+4d)...+a+(n-1)d}
also  S = ( a + ( n − 1 ) d ) . . . + ( a + d ) + a {\displaystyle S=(a+(n-1)d)...+(a+d)+a}
Adding these we get
2 S = ( 2 a + ( n − 1 ) d ) + ( 2 a + ( n − 1 ) d ) + ( 2 a + ( n − 1 ) d ) . . . {\displaystyle 2S=(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)...} n times
Therefore  S = n ( 2 a + ( n − 1 ) d ) / 2 {\displaystyle S=n(2a+(n-1)d)/2}

## Geometric progressions

[edit | edit source]

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

[edit | edit source]

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