# Maths - Progressions

A sequence of 'numbers' or 'quantities' connected in some definite way is called a SERIES and the 'quantity' connected is called a TERM. This page covers:

• Arithmetic Progressions (AP)
• Geometric Progressions (GP)
• Sum to Infinity
• The Binomial Theorem
• Pascals Triangle

## Arithmetic Progressions (AP)

An AP is a series in which each term is formed from the proceeding term by the addition or subtraction of a constant quantity known as the common difference (d).

#### Examples

 2,4,6,8,10 d = 2 16,14,12,10 d = -2 4,0,-4 d = -4

### Equation for the nth term

If we let:

• a = first term
• n = number of terms

the sequence is:

a, a+d, a+2d, a+3d...

So the nth term is:

a+(n-1)d

#### Example

Find the 10th term of 2, 5, 8, 11

10th term = 2+(9*3)

= 29

### To find the sum to n terms of an AP

Let Sn = Sum of 'n' terms, then,

Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ a+(n-1)d

Let l = last term, then,

Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ (l-3d) + (l -2d) + (l - d) + l

reversing the order of this gives:

Sn = l + (l-d) + (l-2d) + (l-3d) + ...+ (a+3d) + (a +2d) + (a + d) + a

Adding these last two equations gives,

2*Sn = a+l + (a+l) + (a+l) + (a+l) + ...+ (a+l) + (a+l) + (a+l) + a+l

2*Sn = (a+l) to n terms

2*Sn = (a+l)*n

Sn = n/2*(a+l)

but l = a+(n-1)d, so substituting this gives:

Sn = n/2*(2a+(n-1)d)

### Arithmetic Mean

When 3 quantities are in arithmetic progression the middle term is the arithmetic mean of the other two.

Find the arithmetic mean of a & b:

Let A be the arithmetic mean, so,

A - a = b - A

2A = a + b

A = (a+b)/2

## Geometric Progressions (GP)

A series of terms, each of which is formed by multiplying the term which proceeds it by a constant factor, known as the common ratio (r)

#### Examples

 3, 6, 12, 24, 48 r = 2 50, 25, 12.5, 6.25 r = 1/2 +9,-27,+81 r = -3

### Equation for the nth term

If we let:

• a = first term
• n = number of terms

the sequence is:

a, ar, ar2, ar3...

So the nth term is:

ar(n-1)

#### Example

Find the 5th term of a GP whose 4th term is 5 and whose 7th term is 320

First find a & r

4th term: ar3 = 5

7th term: ar6 = 320

dividing the 7th term by the 4th term gives:

r3 = 64

therefore:

a = 5/64

r = 3√64 = 4

5th term: ar4 = 5/64 * 44 = 20

### To find the sum to n terms of an GP

Let Sn = Sum of 'n' terms of a series in a GP whose 1stterm is 'a' and common ratio is r.

Sn = a + ar + ar2 + ... + ar(n-2) + ar(n-1)

multiply both sides by r:

r*Sn = ar + ar2 + ar3+ ... + ar(n-1) + arn

subtract the last equation from the one before it:

Sn - r*Sn = a - arn

Sn (1- r)= a (1 - rn)

## Sum to Infinity

It can be shown that:

S = a/(1-r)

#### Example 1

Find the sum to infinity of: 5,-1,1/5...

a = 5

r = -1/5

S = 5/(1-(-1/5)) = 5/(6/5) = 25/6

#### Example 2

The recurring decimal 0.33333 can be written as a series:

0.3 + 0.03 + 0.003 + ...

By finding the sum to infinity express as a fraction:

S = a/(1-r) = 0.3 /(1-1/10) = 0.3/(9/10) = 3/9 = 1/3

## The Binomial Theorem

A binomial expression is a algebraic expression consisting of the sum of the two terms e.g. (a+b), (x-y), (4x2+2y2).

In general any such expression can be denoted by (a+b).

Expansions of (a+b):

(a+b)2= a2+ 2ab + b2

(a+b)3= a3+ 3a2b + 3ab2 + b3

(a+b)4= a4+ 4a3b + 3a2b2+ 4ab3 + b4

It can be seen from the above that the coefficients follow a rule: to find the coefficient for each higher power we add the first and second to get a new second, the second and third to get a new third, the third and fourth to get a new forth and so on.

## Pascals Triangle

(a+b)n= an+ n a(n-1)b + n (n-1)/2! a(n-2) b2+ n (n-1)(n-2)/3! a(n-3)b3 + ... + bn

 n 2n= 0 1 1 1 1 1 1+1 2 1 2 1 1+2+1 3 1 3 3 1 1+3+3+1 4 1 4 6 4 1 1+4+6+4+1 5 1 5 10 10 5 1 1+5+10+10+5+1 6 1 6 15 20 15 6 1 1+6+15+20+15+6+1 7 1 7 21 35 35 21 7 1 1+7+21+35+35+21+7+1 8 1 8 28 56 70 56 28 8 1 1+8+28+56+70+56+28+8+1

when a=1 and b=x the theorem becomes

(1+x)n= 1 + n x + n (n-1)/2! x 2+ n (n-1)(n-2)/3! x 3 + ... + x n

#### example

Expand (2x+3)5

(2x+3)5= (2x)5+ 5(2x)4*3 + (5*4)/(1*2)(2x)3(3)2+(5*4*3)/(1*2*3)(2x)2(3)3+(5*4*3*2)/(1*2*3*4)(2x)(3)4+(5*4*3*2*1)/(1*2*3*4*4)(3)5

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