In the previous section we saw how to determine the nth term of a sequence. We also saw a solved example. Now we will see another solved example. Later in this section we will see how this 'discussion about sequences in general' can be related to Arithmetic progressions.
Solved example 26.37
Fig.26.6 shows a series of squares. The first square have side 1 cm, The second square have side 11⁄2 cm, The third square have side 2 cm, so on.
Write the algebraic expressions for the nth term of the following sequences: (i) Side (ii) Perimeter (iii) Area (iv) Diagonal
Solution:
Part (i), Side:
1. Side of 1st square = 1, Side of 2nd square = 11⁄2, Side of 3rd square = 2, so on
■ So the sequence is: 1, 11⁄2, 2, 21⁄2, 3, 31⁄2, . . .
⇒1, 3⁄2, 2, 5⁄2, 3, 7⁄2, . . .⇒1⁄1, 3⁄2, 2⁄1, 5⁄2, 3⁄1, 7⁄2, . . .
⇒ 2⁄2, 3⁄2, 4⁄2, 5⁄2, 6⁄2, 7⁄2, . . . [Multiplying both numerator and denominator by 2, in those terms which have denominator 1. This will give all terms with denominator 2]
2. Now we see that the numerators form a sequence: 2, 3, 4, 5, . . .
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 more than the 'position number'] of that term
4. So the numerator of the nth term will be (n+1)
5. So the an of the sequence in (1) is given by: an = (n+1)⁄2.
Part (ii), Perimeter:
1. Perimeter of 1st square = 4×1= 4, Perimeter of 2nd square = 4×11⁄2 = 6, Perimeter of 3rd square = 4×2 =8, Perimeter of 4th square = 4×21⁄2 = 10, so on
■ So the sequence is: 4, 6, 8, 10, . . .
2. Let us write the two tasks and their results:
• First task: To find the rule
Result: It is a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [2 more than twice the 'position number'] of that term
• So the algebraic expression for the nth term is: an = 2n+2 = 2(n+1).
Part (iii), Area:
1. Area of 1st square = 1 × 1= 1, Area of 2nd square = 11⁄2 × 11⁄2 = 2.25, Area of 3rd square = 2 × 2 = 4, Area of 4th square = 21⁄2 × 21⁄2 = 6.25, Area of 5th square = 3 × 3 = 9, Area of 6th square = 31⁄2 × 31⁄2 = 12.25, so on
■ So the sequence is: 1, 2.25, 4, 6.25, . . .
⇒ 1, 21⁄4, 4, 61⁄4, 9, 121⁄4 . . .
⇒1, 9⁄4, 4, 25⁄4, 9, 49⁄4, . . .⇒1⁄1, 9⁄4, 4⁄1, 25⁄4, 9⁄1, 49⁄4, . . .
⇒ 4⁄4, 9⁄4, 16⁄4, 25⁄4, 36⁄4, 49⁄4, . . . [Multiplying both numerator and denominator by 4, in those terms which have denominator 1. This will give all terms with denominator 4]
2. Now we see that the numerators form a sequence: 4, 9, 16, 25, 36, 49, . . .
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form 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 more than the 'position number'] of that term
4. So the numerator of the nth term will be (n+1)2
5. So the an of the sequence in (1) is given by: an = (n+1)2⁄4.
Part (iv), Diagonal:
1. Diagonal of 1st square = √[(1)2 + (1)2] = √[2×(1)2] = √2 × √1 = √2
Diagonal of 2nd square = √[(3⁄2)2 + (3⁄2)2] = √[2×(3⁄2)2] = √2 × √[(3⁄2)2] = √2 × 3⁄2.
Diagonal of 3rd square = √[(2)2 + (2)2] = √[2×(2)2] = √2 × √[(2)2] = √2 × 2.
Diagonal of 4th square = √[(5⁄2)2 + (5⁄2)2] = √[2×(5⁄2)2] = √2 × √[(5⁄2)2] = √2 × 5⁄2.
Diagonal of 5th square = √[(3)2 + (3)2] = √[2×(3)2] = √2 × √[(3)2] = √2 × 3.
■ So the sequence is: √2, √2 × 3⁄2, √2 × 2, √2 × 5⁄2, √2 × 3, . . .
⇒√2 × 2⁄2, √2 × 3⁄2, √2 × 4⁄2, √2 × 5⁄2, √2 × 6⁄2, . . .
2. Now we see that the numerators form a sequence: 2, 3, 4, 5, 6, . . .
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 more than the 'position number'] of that term
4. So the numerator of the nth term will be (n+1)
5. All fractions have denominator 2. Also all fractions are multiplied by √2
So the an of the sequence in (1) is given by: an =√2 × [(n+1)⁄2]
• So we now know how to write the algebraic expression for the nth term of any given sequence.
• Now we will see the reverse. That is., we will be given the algebraic expression for the nth term of the sequence. We must be able to write the sequence. Let us see an example:
Solved example 26.38
Write the first four terms of the sequence whose nth term is given by: an = 2n + 3
Solution:
We have: an = 2n + 3
Putting n = 1, 2, 3, 4, we get:
a1 = 2 × 1 + 3 = 2 + 3 = 5,
a2 = 2 × 2 + 3 = 4 + 3 = 7,
a3 = 2 × 3 + 3 = 6 + 3 = 9,
a4 = 2 × 4 + 3 = 8 + 3 = 11
So the required first four terms are 5, 7, 9 and 11
Solved example 26.39
Write the first four terms of the sequence whose nth term is given by: an = n2 + 2
Solution:
We have: an = n2 + 2
Putting n = 1, 2, 3, 4, we get:
a1 = 1 × 1 + 2 = 1 + 2 = 3,
a2 = 2 × 2 + 2 = 4 + 2 = 6,
a3 = 3 × 3 + 2 = 9 + 2 = 11,
a4 = 4 × 4 + 2 = 16 + 2 = 18,
So the required first four terms are 3, 6, 11 and 18
• Method 1: Write sufficient number of terms, starting from the first term a1. By examining those terms, we will be able to write the algebraic expression for the nth term an
• Method 2: Give the algebraic expression for the nth term of a sequence. Using that, we will be able to write the first few number of terms
■ We saw that, if we know the value of the first term 'a' and the common difference 'd' of an AP, the algebraic expression for the nth term of that AP can be written as: an = a + (n-1)d
• This is the general term of an AP. We may be asked to do different types of problems using this general term. Let us analyse:
In the right side of the algebraic expression an = a + (n-1)d, 'a' and 'd' are constants. That is., their values will not change. But 'n' is the 'position number'. It will change.
We can write:
an = a + (n-1)d ⇒ an = a + nd - d ⇒ an = (a-d) + nd ⇒ an = k + nd
So, if we are given a general term of an A.P, we can straight away identify it's common difference 'd'.
Because, 'd' will be the 'coefficient of n'. Let us see an example:
Solved example 26.40
Show that the sequence defined by an = 3n + 4 is an A.P. Also find it's common difference
Solution:
We have: an = 3n + 4
Putting n = 1, 2, 3, 4, we get:
a1 = 3 × 1 + 4 = 3 + 4 = 7,
a2 = 3 × 2 + 4 = 6 + 4 = 10,
a3 = 3 × 3 + 4 = 9 + 4 = 13,
So the first three terms of the sequence are: 7, 10 and 13. We have to prove that they form an A.P. We have done such problems here. Let us use the same steps:
1. Take the last two terms: ak+1 - ak = 13 - 10 = 3
2. Take the preceding two terms with one term common: ak+1 - ak = 10 - 7 = 3
[In this step we have reached the first term]
3. ak+1 - ak is same in the two tests. So it is an AP
4. The common difference d = 3
■ The coefficient of n in the algebraic expression [an = 3n + 4] is also 3
Solved example 26.41
Check whether the sequence defined by an = 3n2 + 2 is an A.P or not
Solution:
We have: an = 3n2 + 2
Putting n = 1, 2, 3, we get:
a1 = 3 × 1 × 1 + 2 = 3 + 2 = 5,
a2 = 3 × 2 × 2 + 2 = 12 + 2 = 14,
a3 = 3 × 3 × 3 + 2 = 27 + 2 = 29,
So the first three terms of the sequence are: 5, 14 and 29. We have to check whether they form an A.P. or not. We have done such problems here. Let us use the same steps:
1. Take the last two terms: ak+1 - ak = 29 - 14 = 15
2. Take the preceding two terms with one term common: ak+1 - ak = 14 - 5 = 9
[In this step we have reached the first term]
3. ak+1 - ak is not same in the two tests. So it is not an AP
This can be proved with out using actual values also. The steps are as follows:
1. We have: an = 3n2 + 2
2. The term after the nth term, that is., the (n+1)th term can be obtained by replacing n with (n+1)
So we get: a(n+1) = 3(n+1)2 + 2 = 3 × (n2 + 2n +1) + 2 = 3n2 + 6n + 3 + 2 = 3n2 + 6n + 5
3. The common difference d = an+1 - an = 3n2 + 6n + 5 - (3n2 + 2) = 3n2 + 6n + 5 - 3n2 - 2 = 6n + 3
4. (6n+3) is not independent of n. That means it will vary with the position of the term. So it is not an A.P
■ In general, if the algebraic expression for the nth term is a linear expression in n, it will be an A.P
We have completed the present discussion on Arithmetic progressions. In the next chapter, we will see Angles inside circles.
Solved example 26.37
Fig.26.6 shows a series of squares. The first square have side 1 cm, The second square have side 11⁄2 cm, The third square have side 2 cm, so on.
 |
| Fig.26.6 |
Solution:
Part (i), Side:
1. Side of 1st square = 1, Side of 2nd square = 11⁄2, Side of 3rd square = 2, so on
■ So the sequence is: 1, 11⁄2, 2, 21⁄2, 3, 31⁄2, . . .
⇒1, 3⁄2, 2, 5⁄2, 3, 7⁄2, . . .⇒1⁄1, 3⁄2, 2⁄1, 5⁄2, 3⁄1, 7⁄2, . . .
⇒ 2⁄2, 3⁄2, 4⁄2, 5⁄2, 6⁄2, 7⁄2, . . . [Multiplying both numerator and denominator by 2, in those terms which have denominator 1. This will give all terms with denominator 2]
2. Now we see that the numerators form a sequence: 2, 3, 4, 5, . . .
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 more than the 'position number'] of that term
4. So the numerator of the nth term will be (n+1)
5. So the an of the sequence in (1) is given by: an = (n+1)⁄2.
Part (ii), Perimeter:
1. Perimeter of 1st square = 4×1= 4, Perimeter of 2nd square = 4×11⁄2 = 6, Perimeter of 3rd square = 4×2 =8, Perimeter of 4th square = 4×21⁄2 = 10, so on
■ So the sequence is: 4, 6, 8, 10, . . .
2. Let us write the two tasks and their results:
• First task: To find the rule
Result: It is a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [2 more than twice the 'position number'] of that term
• So the algebraic expression for the nth term is: an = 2n+2 = 2(n+1).
Part (iii), Area:
1. Area of 1st square = 1 × 1= 1, Area of 2nd square = 11⁄2 × 11⁄2 = 2.25, Area of 3rd square = 2 × 2 = 4, Area of 4th square = 21⁄2 × 21⁄2 = 6.25, Area of 5th square = 3 × 3 = 9, Area of 6th square = 31⁄2 × 31⁄2 = 12.25, so on
■ So the sequence is: 1, 2.25, 4, 6.25, . . .
⇒ 1, 21⁄4, 4, 61⁄4, 9, 121⁄4 . . .
⇒1, 9⁄4, 4, 25⁄4, 9, 49⁄4, . . .⇒1⁄1, 9⁄4, 4⁄1, 25⁄4, 9⁄1, 49⁄4, . . .
⇒ 4⁄4, 9⁄4, 16⁄4, 25⁄4, 36⁄4, 49⁄4, . . . [Multiplying both numerator and denominator by 4, in those terms which have denominator 1. This will give all terms with denominator 4]
2. Now we see that the numerators form a sequence: 4, 9, 16, 25, 36, 49, . . .
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form 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 more than the 'position number'] of that term
4. So the numerator of the nth term will be (n+1)2
5. So the an of the sequence in (1) is given by: an = (n+1)2⁄4.
Part (iv), Diagonal:
1. Diagonal of 1st square = √[(1)2 + (1)2] = √[2×(1)2] = √2 × √1 = √2
Diagonal of 2nd square = √[(3⁄2)2 + (3⁄2)2] = √[2×(3⁄2)2] = √2 × √[(3⁄2)2] = √2 × 3⁄2.
Diagonal of 3rd square = √[(2)2 + (2)2] = √[2×(2)2] = √2 × √[(2)2] = √2 × 2.
Diagonal of 4th square = √[(5⁄2)2 + (5⁄2)2] = √[2×(5⁄2)2] = √2 × √[(5⁄2)2] = √2 × 5⁄2.
Diagonal of 5th square = √[(3)2 + (3)2] = √[2×(3)2] = √2 × √[(3)2] = √2 × 3.
■ So the sequence is: √2, √2 × 3⁄2, √2 × 2, √2 × 5⁄2, √2 × 3, . . .
⇒√2 × 2⁄2, √2 × 3⁄2, √2 × 4⁄2, √2 × 5⁄2, √2 × 6⁄2, . . .
2. Now we see that the numerators form a sequence: 2, 3, 4, 5, 6, . . .
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 more than the 'position number'] of that term
4. So the numerator of the nth term will be (n+1)
5. All fractions have denominator 2. Also all fractions are multiplied by √2
So the an of the sequence in (1) is given by: an =√2 × [(n+1)⁄2]
• So we now know how to write the algebraic expression for the nth term of any given sequence.
• Now we will see the reverse. That is., we will be given the algebraic expression for the nth term of the sequence. We must be able to write the sequence. Let us see an example:
Solved example 26.38
Write the first four terms of the sequence whose nth term is given by: an = 2n + 3
Solution:
We have: an = 2n + 3
Putting n = 1, 2, 3, 4, we get:
a1 = 2 × 1 + 3 = 2 + 3 = 5,
a2 = 2 × 2 + 3 = 4 + 3 = 7,
a3 = 2 × 3 + 3 = 6 + 3 = 9,
a4 = 2 × 4 + 3 = 8 + 3 = 11
So the required first four terms are 5, 7, 9 and 11
Solved example 26.39
Write the first four terms of the sequence whose nth term is given by: an = n2 + 2
Solution:
We have: an = n2 + 2
Putting n = 1, 2, 3, 4, we get:
a1 = 1 × 1 + 2 = 1 + 2 = 3,
a2 = 2 × 2 + 2 = 4 + 2 = 6,
a3 = 3 × 3 + 2 = 9 + 2 = 11,
a4 = 4 × 4 + 2 = 16 + 2 = 18,
So the required first four terms are 3, 6, 11 and 18
From the discussion that we had so far, we can make the following conclusion:
■ A sequence can be specified by two methods:• Method 1: Write sufficient number of terms, starting from the first term a1. By examining those terms, we will be able to write the algebraic expression for the nth term an
• Method 2: Give the algebraic expression for the nth term of a sequence. Using that, we will be able to write the first few number of terms
General term of an Arithmetic progression
We discussed Arithmetic progression in detail earlier in the first section of this chapter.■ We saw that, if we know the value of the first term 'a' and the common difference 'd' of an AP, the algebraic expression for the nth term of that AP can be written as: an = a + (n-1)d
• This is the general term of an AP. We may be asked to do different types of problems using this general term. Let us analyse:
In the right side of the algebraic expression an = a + (n-1)d, 'a' and 'd' are constants. That is., their values will not change. But 'n' is the 'position number'. It will change.
We can write:
an = a + (n-1)d ⇒ an = a + nd - d ⇒ an = (a-d) + nd ⇒ an = k + nd
So, if we are given a general term of an A.P, we can straight away identify it's common difference 'd'.
Because, 'd' will be the 'coefficient of n'. Let us see an example:
Solved example 26.40
Show that the sequence defined by an = 3n + 4 is an A.P. Also find it's common difference
Solution:
We have: an = 3n + 4
Putting n = 1, 2, 3, 4, we get:
a1 = 3 × 1 + 4 = 3 + 4 = 7,
a2 = 3 × 2 + 4 = 6 + 4 = 10,
a3 = 3 × 3 + 4 = 9 + 4 = 13,
So the first three terms of the sequence are: 7, 10 and 13. We have to prove that they form an A.P. We have done such problems here. Let us use the same steps:
1. Take the last two terms: ak+1 - ak = 13 - 10 = 3
2. Take the preceding two terms with one term common: ak+1 - ak = 10 - 7 = 3
[In this step we have reached the first term]
3. ak+1 - ak is same in the two tests. So it is an AP
4. The common difference d = 3
■ The coefficient of n in the algebraic expression [an = 3n + 4] is also 3
Solved example 26.41
Check whether the sequence defined by an = 3n2 + 2 is an A.P or not
Solution:
We have: an = 3n2 + 2
Putting n = 1, 2, 3, we get:
a1 = 3 × 1 × 1 + 2 = 3 + 2 = 5,
a2 = 3 × 2 × 2 + 2 = 12 + 2 = 14,
a3 = 3 × 3 × 3 + 2 = 27 + 2 = 29,
So the first three terms of the sequence are: 5, 14 and 29. We have to check whether they form an A.P. or not. We have done such problems here. Let us use the same steps:
1. Take the last two terms: ak+1 - ak = 29 - 14 = 15
2. Take the preceding two terms with one term common: ak+1 - ak = 14 - 5 = 9
[In this step we have reached the first term]
3. ak+1 - ak is not same in the two tests. So it is not an AP
This can be proved with out using actual values also. The steps are as follows:
1. We have: an = 3n2 + 2
2. The term after the nth term, that is., the (n+1)th term can be obtained by replacing n with (n+1)
So we get: a(n+1) = 3(n+1)2 + 2 = 3 × (n2 + 2n +1) + 2 = 3n2 + 6n + 3 + 2 = 3n2 + 6n + 5
3. The common difference d = an+1 - an = 3n2 + 6n + 5 - (3n2 + 2) = 3n2 + 6n + 5 - 3n2 - 2 = 6n + 3
4. (6n+3) is not independent of n. That means it will vary with the position of the term. So it is not an A.P
■ In general, if the algebraic expression for the nth term is a linear expression in n, it will be an A.P
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