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,
  2. Test for AP or GP

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


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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