Definitions Arithmetic Progression Geometric Progression General Examples

efinitions

1. A sequence is a set of terms in a defined order with a rule for forming the terms.
2. A series is the sum of the terms of a sequence.
3. A progression can be a sequence or a series.

rithmetic Progression

An arithemtic progression (AP) is a sequence of terms in which any term minus its immediate preceding term gives a constant.

This constant is called the common difference.

Let a be the first term and d be the common difference of an AP.

The nth term of the AP is  T_n = a + (n - 1)d.

The sum of the first n terms is Sn	= ½n[2a + (n - 1)d]
	= ½n[a + Tn].

eometric Progression

A geometric progression (GP) is a sequence in which each term is obtained from the preceding one by multplying it by a constant.

This constant is called the common ratio.

Let a be the first term and r be the common ratio of a GP.

The nth term of the AP is T_n = ar^{n - 1}.

The sum of the first n terms is Sn =

a(rn - 1) ¾¾¾¾ r - 1

or

a(1 - rn) ¾¾¾¾ 1 - r

.

The sum to infinity is S¥ =

a ¾¾¾ 1 - r

provided that |r| < 1.

eneral

1. For any progression, if S_n is given,
   • the first term of the progression is given by T₁ = S₁.
   • The nth term is given by T_n = S_n - S_{n - 1}.
2. Test for AP or GP

   Suppose we know the formula for general term T_n of a progression.
   Compute the term T_{n + 1}.

   • If the difference T_{n + 1} - T_n is a constant, the progression is an AP.
   • If the ratio T_{n + 1}/T_n is a constant, the progression is a GP.

xamples

Example 1:  Find the sum of all integers between 1000 and 3000 inclusive which are not divisible by 7.

Solution:

n = 3000 - 1000 + 1 = 2001

\	Sum of all integers between 1000 and 3000
	= (2001/2)[1000 + 3000]
	= 4 002 000

Numbers between 1000 and 3000 divisible by 7 are:

1001, 1008, ..., 2996.

This is an A.P. with a = 1001 and d = 7.

2996 = 1001 + (n - 1)7  Þ  n = 286.

\	Sum of all integers between 1000 and 3000 divisible by 7
	= (286/2)[1001 + 2996]
	= 571 571

\	Sum of all integers between 1000 and 3000 which are not divisible by 7
	= 4 002 000 - 571 571
	= 3 430 429

Example 2:  An arithmetic series has first term 2000 and common difference -2.3.  Calculate the value of the first negative term of the series, and the sum of all the positive terms.

Solution:

Tn = a + (n - 1)d	<	0
2000 + (n - 1)(-2.3)	<	0
(n - 1)(-2.3)	<	-2000
n - 1	>	-2000/-2.3
n	>	870.56...

The first negative term is the 871st term.

\  Value of first negative term	= 2000 + (871 - 1)(-2.3)
	= -1

Number of positive terms = 870.

\  Sum of all positive terms	= (870/2)[4000 + (870 - 1)(-2.3)]
	= 1 741 131

Example 3:  The first two terms of a G.P. are 4 and -3.  Write down an expression for the sum to n terms and evaluate the sum to infinity of this progression.  Find the least value of n for the sum to n terms to be within 1% of the sum to infinity.

Solution: