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 n^th term

If we let:

• a = first term
• n = number of terms

the sequence is:

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

So the n^th term is:

a+(n-1)d

Example

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

10^th term = 2+(9*3)

= 29

To find the sum to n terms of an AP

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

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

Let l = last term, then,

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

reversing the order of this gives:

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

Adding these last two equations gives,

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

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

2*S_n = (a+l)*n

S_n = n/2*(a+l)

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

S_n = 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 n^th term

If we let:

• a = first term
• n = number of terms

the sequence is:

a, ar, ar², ar³...

So the n^th term is:

ar^(n-1)

Example

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

First find a & r

4^th term: ar³ = 5

7^th term: ar⁶ = 320

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

r³ = 64

therefore:

a = 5/64

r = ³√64 = 4

5^th term: ar⁴ = 5/64 * 4⁴ = 20

To find the sum to n terms of an GP

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

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

multiply both sides by r:

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

subtract the last equation from the one before it:

S_n - r*S_n = a - arⁿ

S_n (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

Further Reading

For information about infinite series and there use to calculate the value of functions see this page.

Elements of Equations

Trigonometry Functions