24 Arithmetic Progressions
By studying this lesson, you will be able to
recognize arithmetic progressions and solve problems related to arithmetic
progressions.
In earlier grades, you have learned various number patterns. A number pattern
when indicated as a list is called a sequence of numbers or simply a sequence. Let
us consider the following sequence.
3, 8, 13, 18, ....
In this sequence, the first term is 3, the second term is 8, the third term is 13 etc.
A feature of this sequence is that, considering any two consecutive terms in the
sequence, when the first term is subtracted from the second term, a constant value
is obtained. In this case, the constant value is 5.
A similar sequence is shown below.
8, 5, 2, –1" –4, '''
In this sequence too, when the first term is subtracted from the second term, for any
pair of consecutive terms, a constant value is obtained. In this case, the constant
value is –3.
Sequences having this feature are called arithmetic progressions. The constant
value obtained by subtracting any term from the term right after that term is called
the common difference, and is usually denoted by d.
An arithmetic progression is a sequence of numbers such that
a constant value is obtained when any term is subtracted from
the term right after that term.
The common difference d, of an arithmetic progression can be found as follows:
common difference (d) = (any term other than the first term) – (the preceding term)
For free distribution 61
, Review Exercise
1. Determine whether each of the following sequences is an arithmetic progression.
2. Find the common difference of each of the following arithmetic progressions.
24.1 nth term of an Arithmetic Progression
The following notation is used to denote the terms of an arithmetic progression.
T1= 1st term
T2= 2nd term
T3= 3rd term etc.
For example, for the arithmetic progression 6, 8, 10, 12, 14,...
we may write T1= 6" T2= 8, T3= 10, T4= 12, T5= 14 etc .
What is the 25th term of this progression? In other words, what is the value of T25?
It is clear that if you continue writing the terms according to the above pattern, the
25th term appears when you write 25 terms. If you do this, you will get 54 as the 25th
term. That is, T25 = 54.
Now, if you require to find the 500th term of this progression, how would you find
it? For this you would have to write down 500 terms following the pattern, which
is quite a tedious task. Let us see how we can derive a formula that can be used to
62 For free distribution
, find any term of an arithmetic progression rather easily.
Let us illustrate this derivation using the above arithmetic progression 6, 8, 10, 12,
... For this progression, the first term is 6 and the common difference is 2. Observe
carefully how the terms of this progression have been written in terms of the first
term and the common difference in the following table.
Term Value of the Value of the term in terms of the first
term term and the common difference
T1 6 6 = 6 + (1 – 1) × 2
T2 8 6+2 = 6 + (2 – 1) × 2
T3 10 6+2+2 = 6 + (3 – 1) × 2
T4 12 6 + 2 + 2 + 2 = 6 + (4 – 1) × 2
... ''' ... ...
Now, according to the pattern in the table,
T500 = 6 + (500 – 1) × 2
= 6 + 499 × 2
= 6 + 998
= 1004
Hence, the 500th term is 1004.
Can you generalize the above pattern further? In other words, can you find a formula
for the nth term Tn in terms of the first term a and the common difference d? For
this, look at the expression T500 = 6 + (500 – 1) × 2 again, where 6 is the 1st term and
2 is the common difference.
You can see that if you follow the above pattern to obtain the nth term Tn of the
arithmetic progression with first term a and common difference d, you obtain
Tn = a + (n – 1) d. In this formula, according to our notation, Tn denotes the nth term.
Hence, the nth term Tn of the arithmetic progression with first term a and common
difference d is given by
Tn = a + (n – 1) d
The importance of this formula is that it gives the relationship between the four
unknowns a, d, n and Tn. This formula can be used to find the value of any one
of the four unknowns when the values of the other 3 unknowns in an arithmetic
progression are known.
Now let us consider how problems on arithmetic progressions are solved using this
formula.
For free distribution 63
By studying this lesson, you will be able to
recognize arithmetic progressions and solve problems related to arithmetic
progressions.
In earlier grades, you have learned various number patterns. A number pattern
when indicated as a list is called a sequence of numbers or simply a sequence. Let
us consider the following sequence.
3, 8, 13, 18, ....
In this sequence, the first term is 3, the second term is 8, the third term is 13 etc.
A feature of this sequence is that, considering any two consecutive terms in the
sequence, when the first term is subtracted from the second term, a constant value
is obtained. In this case, the constant value is 5.
A similar sequence is shown below.
8, 5, 2, –1" –4, '''
In this sequence too, when the first term is subtracted from the second term, for any
pair of consecutive terms, a constant value is obtained. In this case, the constant
value is –3.
Sequences having this feature are called arithmetic progressions. The constant
value obtained by subtracting any term from the term right after that term is called
the common difference, and is usually denoted by d.
An arithmetic progression is a sequence of numbers such that
a constant value is obtained when any term is subtracted from
the term right after that term.
The common difference d, of an arithmetic progression can be found as follows:
common difference (d) = (any term other than the first term) – (the preceding term)
For free distribution 61
, Review Exercise
1. Determine whether each of the following sequences is an arithmetic progression.
2. Find the common difference of each of the following arithmetic progressions.
24.1 nth term of an Arithmetic Progression
The following notation is used to denote the terms of an arithmetic progression.
T1= 1st term
T2= 2nd term
T3= 3rd term etc.
For example, for the arithmetic progression 6, 8, 10, 12, 14,...
we may write T1= 6" T2= 8, T3= 10, T4= 12, T5= 14 etc .
What is the 25th term of this progression? In other words, what is the value of T25?
It is clear that if you continue writing the terms according to the above pattern, the
25th term appears when you write 25 terms. If you do this, you will get 54 as the 25th
term. That is, T25 = 54.
Now, if you require to find the 500th term of this progression, how would you find
it? For this you would have to write down 500 terms following the pattern, which
is quite a tedious task. Let us see how we can derive a formula that can be used to
62 For free distribution
, find any term of an arithmetic progression rather easily.
Let us illustrate this derivation using the above arithmetic progression 6, 8, 10, 12,
... For this progression, the first term is 6 and the common difference is 2. Observe
carefully how the terms of this progression have been written in terms of the first
term and the common difference in the following table.
Term Value of the Value of the term in terms of the first
term term and the common difference
T1 6 6 = 6 + (1 – 1) × 2
T2 8 6+2 = 6 + (2 – 1) × 2
T3 10 6+2+2 = 6 + (3 – 1) × 2
T4 12 6 + 2 + 2 + 2 = 6 + (4 – 1) × 2
... ''' ... ...
Now, according to the pattern in the table,
T500 = 6 + (500 – 1) × 2
= 6 + 499 × 2
= 6 + 998
= 1004
Hence, the 500th term is 1004.
Can you generalize the above pattern further? In other words, can you find a formula
for the nth term Tn in terms of the first term a and the common difference d? For
this, look at the expression T500 = 6 + (500 – 1) × 2 again, where 6 is the 1st term and
2 is the common difference.
You can see that if you follow the above pattern to obtain the nth term Tn of the
arithmetic progression with first term a and common difference d, you obtain
Tn = a + (n – 1) d. In this formula, according to our notation, Tn denotes the nth term.
Hence, the nth term Tn of the arithmetic progression with first term a and common
difference d is given by
Tn = a + (n – 1) d
The importance of this formula is that it gives the relationship between the four
unknowns a, d, n and Tn. This formula can be used to find the value of any one
of the four unknowns when the values of the other 3 unknowns in an arithmetic
progression are known.
Now let us consider how problems on arithmetic progressions are solved using this
formula.
For free distribution 63
