4 Jan 1999

### 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  Tn = 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 Tn = arn - 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 Sn is given,
• the first term of the progression is given by T1 = S1.
• The nth term is given by Tn = Sn - Sn - 1.
2. Test for AP or GP

Suppose we know the formula for general term Tn of a progression.
Compute the term Tn + 1.

• If the difference Tn + 1 - Tn is a constant, the progression is an AP.
• If the ratio Tn + 1/Tn 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:

Sum to n terms, Sn =
4[1 - (-3/4)n]
¾¾¾¾¾¾
1 - (-3/4)

=
16[1 - (-3/4)n]
¾¾¾¾¾¾
7

Sum to infinity, S¥ =
 4 ¾¾¾¾ 1 - (-3/4)

= 16/7

|Sn - S¥|
<  1%  S¥
|(16/7)[1 - (-3/4)n] - 16/7|
<  (1/100)(16/7)
(16/7)|(-3/4)n|
<  (1/100)(16/7)
(3/4)n
<  1/100
n lg (3/4)
<  lg (1/100)

n
>
 lg (1/100) ¾¾¾¾ lg (3/4) = 16.0...

\  Least value of n = 17.

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