Tuesday, April 25, 2017

Chapter 26.6 - Basic details about Sequences

In the previous section we completed the discussion on Arithmetic progressionsIn this section we will see sequences in general.

• Consider the following arrangement of numbers:
1, 4, 9, 16, . . .
• In this arrangement, the numbers are arranged according to a rule. We must do the following two tasks:
1. Find the rule
2. Find how the rule is applied

1. Finding the rule:
• If we carefully examine the numbers, the rule can be found out without much difficulty. We find that the numbers are all squares.
    ♦ 1 is the square of 1 
    ♦ 4 is the square of 2
    ♦ 9 is the square of 3
so on
• So the rule is: Squares of natural numbers are arranged in a sequence
2. Finding how the rule is applied:
• For this task, we have to first familiarise ourselves with some general procedures associated with sequences. Let us see them:
(a) The various numbers occurring in the sequence are called terms
(b) The terms are denoted as a1, a2, a3, . . . etc., OR x1, x2, x3, . . . etc.,
(c) The subscripts denote the position of the terms. 
    ♦ a1 is the 1st term
    ♦ ais the 2nd term
    ♦ a3 is the 3rd  term
so on
■ So subscripts are the 'position numbers'
Now let us proceed with our second task:
• For our given sequence, we can write:
a1 = 12, a2 = 22, a3 = 32 so on
■ From the above step, we can note one point:
Consider any position. The actual term at that position will be the square of the 'position number'. For example:
    ♦ awill be 52.
    ♦ awill be 92.
■ So, if we consider the nth term,
• It is denoted as an
• It's 'position number' is n
• The actual term at that nth position is n2.
■ So we can write: an2.
Thus our second task is also complete. Let us summarise:
• First task: To find the rule
Result: It is a sequence of squares of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the square of the 'position number' of that term. 
• This gives an expressionan2
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.

■ To understand the two tasks and their results, let us consider a variant of the above sequence:
0, 1, 4, 9, 16, . . .
The tasks and their results are:
• First task: To find the rule
Result: It is a sequence of squares of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the square of [1 subtracted from the 'position number'] of that term
• This gives an expressiona= (n-1)2 
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.
• Note that, in the variant, rule is the same as before. But the way in which the rule is applied is different
■ Another variant:
1, 0, 1, 4, 9, 16, . . .
The tasks and their results are:
• First task: To find the rule
Result: It is a sequence of squares of natural numbers
• Second task: To find how the rule is applied
• Result: Any term taken from the sequence is the square of [2 subtracted from the 'position number'] of that term
• This gives an expressiona= (n-2)2 
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.
• Note that, in this variant also, rule is the same as before. But the way in which the rule is applied is different 

■ Now we will see another example. Consider the sequence:
1, 12, 13, 14, 15, . . .
In this sequence, a1 = 11a2 = 12a3 = 13, so on  
Let us write the two tasks and their results:
• First task: To find the rule
Result: It is a sequence of reciprocals of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the reciprocal of the 'position number' of that term
• This gives an expression: an = 1n
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.
■ One more example:
1, 3, 5, 7, 9, . . .
In this sequence, a1 =1, a2 = 3, a3 = 5, so on  
Let us write the two tasks and their results:
• First task: To find the rule
Result: It is a sequence of odd numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 subtracted from twice the 'position number'] of that term
• This gives an expression: a= 2n - 1
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.

So we saw three examples of sequences. In each of them, we performed the two tasks and obtained their results. We can now give the definition for sequences:
■ A sequence is an arrangement of numbers in a definite order according to some rule

In any sequence, as a result of performing the two tasks, we will get an expression for an, which is the nth term.
• In the first example we got: an2   
• In the second example we got: an = 1n
• In the third example we got: a= 2n - 1 
■ This nth term is called the general term of the sequence.

We will now see some solved examples:
Solved example 26.36
One cubic cm of iron weighs 7.8 grams. Write as sequence:
(i) The volumes of iron cubes of sides 1 cm, 2 cm, 3 cm, . . .
(ii) The weights of iron cubes of sides 1 cm, 2 cm, 3 cm, . . .
(iii) Also write the algebraic expression for the nth term of the above sequences
Solution:
Part (i), Volumes:
• Volume of a cube of side 1 cm = 13 = 1 cm3
• Volume of a cube of side 2 cm = 23 = 8 cm3
• Volume of a cube of side 3 cm = 33 = 27 cm3.
■ So the sequence is: 1, 8, 27, . . .
Part (ii), Weights:
• Weight of a cube of side 1 cm = volume × weight per unit volume = 1 × 7.8 = 7.8 grams
• Weight of a cube of side 2 cm = 8 × 7.8 = 62.4 grams
• Weight of a cube of side 3 cm = 27 × 7.8 = 210.6 grams
■ So the sequence is: 7.8, 62.4, 210.6, . . .
Part (iii), Algebraic expressions
1. Consider the first sequence:
1, 8, 27, . . .
In this sequence, a1 =  13 = 1 , a2 = 23 = 8a3 = 33 = 27, so on  
• First task: To find the rule
Result: It is a sequence of cubes of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the cube of the 'position number' of that term. 
• So the algebraic expression for the nth term isan3.
2. Consider the second sequence:
7.8, 62.4, 210.6, . . .
In this sequence, a1 = 7.8 × 13 = 7.8a2 = 7.8 × 23 = 62.4a3 = 7.8 × 33 = 210.4, so on  
• First task: To find the rule
Result: It is a sequence of cubes of natural numbers multiplied by 7.8
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the cube of the 'position number' of that term, multiplied by 7.8  
• So the algebraic expression for the nth term isa= 7.8n3


In the next section we will see another solved example. We will also see how the above discussion is related to Arithmetic progressions.


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