In the previous section we learned how to determine whether a given series is an AP or not. We can also write it's first term and common difference. In this section we will learn to write the nth term of an AP.
Earlier we saw the general form of an AP. It was:
a, a+d, a+2d, a+3d, . . . [Details here]
From this general form, we can write:
• 1st term of the AP = a + 0 × d
• 2nd term of the AP = a + 1 × d
• 3rd term of the AP = a + 2 × d
• 4th term of the AP = a + 3 × d
So on . . .
■ We see a pattern here:
• For the 1st term a1, coefficient of d is (1-1) = 0
• For the 2nd term a2, coefficient of d is (2-1) = 1
• For the 3rd term a3, coefficient of d is (3-1) = 2
• For the 4th term a4, coefficient of d is (4-1) = 3
So on . . .
■ Thus we see a relation between the following two:
• The number of the term
• The coefficient of d
■ The relation is this:
The 'coefficient of d' is '1 less than the number of the term'
Example:
• The coefficient of d in the 16th term will be (16-1) = 15
♦ So the 16th term a16 will be a+15d
• The coefficient of d in the 38th term will be (38-1) = 37
♦ So the 38th term a38 will be a+37d
■ So we can write an equation for general use:
Eq.26.1:
The nth term an of an AP = a + (n-1)d.
Where a is the first term and d is the common difference.
an is also called the general term of the AP. If there are m terms in the AP, then am represents the last term. This last term is sometimes denoted by l.
Now we will see some solved examples.
Solved example 26.6
Find the 10th term of the AP: 2, 7, 12, . . .
Solution:
1. 1st term a = 2
2. d = 12 - 7 = 5 [Since it is given as an AP, we do not need to check d for all terms]
3. We have: nth term = a + (n-1)d
So 10th term = 2 + (10-1) × 5 = 2 + 9 × 5 = 2 + 45 = 47
Solved example 26.7
Which term of the AP : 21, 18, 15, . . . is -81? Also, is any term 0? Give reason for your answer.
Solution:
1. 1st term a = 21
2. d = 15 - 18 = -3 [Since it is given as an AP, we do not need to check d for all terms]
3. We have: nth term = a + (n-1)d
4. Let -81 be the nth term. Then substituting the values in (3) we get:
5. -81 = 21 + (n-1) × -3 ⇒ -81 = 21 -3n + 3 ⇒ 3n = 81 +21 +3 ⇒ 3n = 105 ⇒ n = 35
So -81 is the 35th term of the given AP. This is the answer for part (i)
6. Let 0 be the term. Substituting the values in (3) we get:
7. 0 = 21 + (n-1) × -3 ⇒ 0 = 21 -3n + 3 ⇒ 3n = 21 +3 ⇒ 3n = 24 ⇒ n = 8
8. We get n as a whole number. So 0 is indeed a term in the series. And it is the 8th term
Solved example 26.8
Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) 5 = a + (3-1) × d ⇒ 5 = a + 2d
(ii) 9 = a + (7-1) × d ⇒ 9 = a + 6d
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 5 - 2d
5. Substituting this in (ii) we get: 9 = (5 - 2d) + 6d
⇒ 9 = 5 + 4d ⇒ 4d = 4 ⇒ d = 1
6. Substituting this value of d in (4) we get: a = 5 - 2 × 1 ⇒ a = 5 -2 ⇒ a = 3
7. So we have a and d. We can now form the AP.
1st term = a1 = a = 3
2nd term = a2 = a + d = 3 + 1 = 4
3rd term = a3 = a + 2d = 3 + 2×1 = 3 + 2 = 5
4th term = a4 = a + 3d = 3 + 3×1 = 3 + 3 = 6
Thus the AP is: 3, 4, 5, 6, . . .
Solved example 26.9
Check whether 301 is a term in the series: 5, 11, 17, 23, . . .
Solution:
1. In this problem, it is not given that it is an AP. So we have to first find whether it is an AP or not.
(i) Take the last two terms: ak+1 - ak = 23 - 17 = 6
(ii) Take the preceding two terms with one term common: ak+1 - ak = 17 - 11 = 6
(iii) Take the preceding two terms with one term common: ak+1 - ak = 11 - 5 = 6
[In this step we have reached the first term]
(iv) ak+1 - ak is same in the three tests. So it is an AP
(v) The common difference d = 6
2. 1st term a = 5
3. We have: nth term = a + (n-1)d
4. Let 301 be the nth term. Then substituting the values in (3) we get:
5. 301 = 5 + (n-1) × 6 ⇒ 301 = 5 + 6n - 6 ⇒ 6n = 301 -5 + 6 ⇒ 6n = 302 ⇒ n = 302⁄6 = 151⁄3.
• 'n' is a number. It shows a position. Like 1st, 2nd, 3rd etc., So 'n' should be a whole number.
• 151⁄3 is not a whole number. So 301 is not a term in the given AP.
Solved example 26.10
How many two-digit numbers are divisible by 3?
Solution:
• There are so many 2 digit numbers. But the list is not very long. All the 2 digit numbers fall between 10 and 99 both of them inclusive.
• Out of them, many are divisible by 3. In this problem, we do not have to actually pick out all those which are divisible by 3.
• All we need is the 'number of 2-digit numbers' which are divisible by 3. Let us try to find it:
1. • The first 2-digit number is 10. It is not divisible by 3
• The second 2-digit number is 11. It is not divisible by 3
• The third 2-digit number is 12. It is divisible by 3
• The fourth 2-digit number is 13. It is not divisible by 3
so on . . .
• The last 2-digit number is 99. It is divisible by 3
2. Let us write the list of those 2-digit numbers which are divisible by 3:
12, 15, 18, 21, . . . , 99.
3. This list is an AP. Because we get any term by adding 3 to the preceding term. So we can apply the principles of AP to this list.
4. 1st term a = 12, common difference d = 3
5. So we can write: nth term = a + (n-1)d ⇒ nth term = 12 + (n-1) ×3
6. Let 99 be the nth term. Then we can write:
99 = 12 + (n-1) ×3 ⇒ 99 = 12 + 3n -3 ⇒ 3n = 90 ⇒ n = 30
7. So the last term 99 is the 30th term. We can write:
There are 30 two-digit numbers which are divisible by 3
Solved example 26.11
Find the 11th term from the last term (towards the 1st term) of the AP : 10, 7, 4, . . ., – 62.
Solution:
• In this problem, all details of the AP is given to us. Including the last term.
• We need to find the position of the last term. That position will give us the total number of terms in the AP
1. 1st term a = 10, common difference d = 4 - 7 = -3
2. So we can write: nth term = a + (n-1)d ⇒ nth term = 10 + (n-1) ×-3
3. Let -62 be the nth term. Then we can write:
-62 = 10 + (n-1) ×-3 ⇒ -62 = 10 - 3n +3 ⇒ 3n = 62 + 13 ⇒ 3n = 75 ⇒ n = 25
4. So there are a total of 25 terms in the AP
5. We want the 11th term from the last term. That is., (25 - 11 +1)th term = 15th term from the first
• In fig.a, there are a total of 12 terms. If we want the 5th term from the last, it will be the (12-5+1)th term = 8th term from the first.
■ That means, we subtract the 'required term from the last' from the total and then add 1
• Similarly, in fig.b, If we want the 9th term from the last, it will be the (12-9+1)tth term = 4th term from the first.
Solved example 26.12
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Solution:
1. Let us write all the given values in the form of a series:
23, 21, 19, . . . ,5
2. This is an AP with d = -2. Because every term can be obtained by adding (-2) to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term 5
4. We have: nth term = a + (n-1)d ⇒ 5 = 23 + (n-1) ×-2
⇒ 5 = 23 - 2n + 2 ⇒ 2n = 23 - 5 + 2 ⇒ 2n = 20 ⇒ n = 10
• So there are a total of 10 rows in the flower bed.
Solved example 26.13
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
Solution:
We have: nth term = an = a + (n-1)d
(i) a, d and n are given. an is the unknown. We can use the formula directly.
an = 7 + (8-1)×3 = 7 + 7 × 3 = 7 + 21 = 28
(ii) Here d is the unknown
0 = -18 + (10-1)×d ⇒ 0 = -18 + 9d ⇒ 9d = 18 ⇒ d = 2
(iii) Here a is the unknown
-5 = a + (18-1)×-3 ⇒ -5 = a - 51 ⇒ a = 51 -5 = 46
(iv) Here n is the unknown
3.6 = -18.9 + (n-1)×2.5 ⇒ 3.6 = -18.9 + 2.5n -2.5 ⇒ 3.6 = -21.4 + 2.5n ⇒ 2.5n = 25 ⇒ n = 10
(v) a, d and n are given. an is the unknown. We can use the formula directly.
an = 3.5 + (105-1)×0 = 3.5
• Also note that, since d is zero, all terms will be equal to the first term. So 105th term is also equal to 3.5
Solved example 26.14
Which term of the AP: 3, 8, 13, 18, . . . is 78
Solution:
1. 1st term a = 3, common difference d = 18 - 13 = 5
2. So we can write: nth term = a + (n-1)d ⇒ nth term = 3 + (n-1) ×5
3. Let 78 be the nth term. Then we can write:
78 = 3 + (n-1) ×5 ⇒ 78 = 3 + 5n - 5 ⇒ 5n = 78 + 5 - 3 ⇒ 5n = 80 ⇒ n = 16
In the next section we will see a few more solved examples.
Earlier we saw the general form of an AP. It was:
a, a+d, a+2d, a+3d, . . . [Details here]
From this general form, we can write:
• 1st term of the AP = a + 0 × d
• 2nd term of the AP = a + 1 × d
• 3rd term of the AP = a + 2 × d
• 4th term of the AP = a + 3 × d
So on . . .
■ We see a pattern here:
• For the 1st term a1, coefficient of d is (1-1) = 0
• For the 2nd term a2, coefficient of d is (2-1) = 1
• For the 3rd term a3, coefficient of d is (3-1) = 2
• For the 4th term a4, coefficient of d is (4-1) = 3
So on . . .
■ Thus we see a relation between the following two:
• The number of the term
• The coefficient of d
■ The relation is this:
The 'coefficient of d' is '1 less than the number of the term'
Example:
• The coefficient of d in the 16th term will be (16-1) = 15
♦ So the 16th term a16 will be a+15d
• The coefficient of d in the 38th term will be (38-1) = 37
♦ So the 38th term a38 will be a+37d
■ So we can write an equation for general use:
Eq.26.1:
The nth term an of an AP = a + (n-1)d.
Where a is the first term and d is the common difference.
an is also called the general term of the AP. If there are m terms in the AP, then am represents the last term. This last term is sometimes denoted by l.
Now we will see some solved examples.
Solved example 26.6
Find the 10th term of the AP: 2, 7, 12, . . .
Solution:
1. 1st term a = 2
2. d = 12 - 7 = 5 [Since it is given as an AP, we do not need to check d for all terms]
3. We have: nth term = a + (n-1)d
So 10th term = 2 + (10-1) × 5 = 2 + 9 × 5 = 2 + 45 = 47
Solved example 26.7
Which term of the AP : 21, 18, 15, . . . is -81? Also, is any term 0? Give reason for your answer.
Solution:
1. 1st term a = 21
2. d = 15 - 18 = -3 [Since it is given as an AP, we do not need to check d for all terms]
3. We have: nth term = a + (n-1)d
4. Let -81 be the nth term. Then substituting the values in (3) we get:
5. -81 = 21 + (n-1) × -3 ⇒ -81 = 21 -3n + 3 ⇒ 3n = 81 +21 +3 ⇒ 3n = 105 ⇒ n = 35
So -81 is the 35th term of the given AP. This is the answer for part (i)
6. Let 0 be the term. Substituting the values in (3) we get:
7. 0 = 21 + (n-1) × -3 ⇒ 0 = 21 -3n + 3 ⇒ 3n = 21 +3 ⇒ 3n = 24 ⇒ n = 8
8. We get n as a whole number. So 0 is indeed a term in the series. And it is the 8th term
Solved example 26.8
Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) 5 = a + (3-1) × d ⇒ 5 = a + 2d
(ii) 9 = a + (7-1) × d ⇒ 9 = a + 6d
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 5 - 2d
5. Substituting this in (ii) we get: 9 = (5 - 2d) + 6d
⇒ 9 = 5 + 4d ⇒ 4d = 4 ⇒ d = 1
6. Substituting this value of d in (4) we get: a = 5 - 2 × 1 ⇒ a = 5 -2 ⇒ a = 3
7. So we have a and d. We can now form the AP.
1st term = a1 = a = 3
2nd term = a2 = a + d = 3 + 1 = 4
3rd term = a3 = a + 2d = 3 + 2×1 = 3 + 2 = 5
4th term = a4 = a + 3d = 3 + 3×1 = 3 + 3 = 6
Thus the AP is: 3, 4, 5, 6, . . .
Solved example 26.9
Check whether 301 is a term in the series: 5, 11, 17, 23, . . .
Solution:
1. In this problem, it is not given that it is an AP. So we have to first find whether it is an AP or not.
(i) Take the last two terms: ak+1 - ak = 23 - 17 = 6
(ii) Take the preceding two terms with one term common: ak+1 - ak = 17 - 11 = 6
(iii) Take the preceding two terms with one term common: ak+1 - ak = 11 - 5 = 6
[In this step we have reached the first term]
(iv) ak+1 - ak is same in the three tests. So it is an AP
(v) The common difference d = 6
2. 1st term a = 5
3. We have: nth term = a + (n-1)d
4. Let 301 be the nth term. Then substituting the values in (3) we get:
5. 301 = 5 + (n-1) × 6 ⇒ 301 = 5 + 6n - 6 ⇒ 6n = 301 -5 + 6 ⇒ 6n = 302 ⇒ n = 302⁄6 = 151⁄3.
• 'n' is a number. It shows a position. Like 1st, 2nd, 3rd etc., So 'n' should be a whole number.
• 151⁄3 is not a whole number. So 301 is not a term in the given AP.
Solved example 26.10
How many two-digit numbers are divisible by 3?
Solution:
• There are so many 2 digit numbers. But the list is not very long. All the 2 digit numbers fall between 10 and 99 both of them inclusive.
• Out of them, many are divisible by 3. In this problem, we do not have to actually pick out all those which are divisible by 3.
• All we need is the 'number of 2-digit numbers' which are divisible by 3. Let us try to find it:
1. • The first 2-digit number is 10. It is not divisible by 3
• The second 2-digit number is 11. It is not divisible by 3
• The third 2-digit number is 12. It is divisible by 3
• The fourth 2-digit number is 13. It is not divisible by 3
so on . . .
• The last 2-digit number is 99. It is divisible by 3
2. Let us write the list of those 2-digit numbers which are divisible by 3:
12, 15, 18, 21, . . . , 99.
3. This list is an AP. Because we get any term by adding 3 to the preceding term. So we can apply the principles of AP to this list.
4. 1st term a = 12, common difference d = 3
5. So we can write: nth term = a + (n-1)d ⇒ nth term = 12 + (n-1) ×3
6. Let 99 be the nth term. Then we can write:
99 = 12 + (n-1) ×3 ⇒ 99 = 12 + 3n -3 ⇒ 3n = 90 ⇒ n = 30
7. So the last term 99 is the 30th term. We can write:
There are 30 two-digit numbers which are divisible by 3
Solved example 26.11
Find the 11th term from the last term (towards the 1st term) of the AP : 10, 7, 4, . . ., – 62.
Solution:
• In this problem, all details of the AP is given to us. Including the last term.
• We need to find the position of the last term. That position will give us the total number of terms in the AP
1. 1st term a = 10, common difference d = 4 - 7 = -3
2. So we can write: nth term = a + (n-1)d ⇒ nth term = 10 + (n-1) ×-3
3. Let -62 be the nth term. Then we can write:
-62 = 10 + (n-1) ×-3 ⇒ -62 = 10 - 3n +3 ⇒ 3n = 62 + 13 ⇒ 3n = 75 ⇒ n = 25
4. So there are a total of 25 terms in the AP
5. We want the 11th term from the last term. That is., (25 - 11 +1)th term = 15th term from the first
This step will be clear from the fig.26.2 below:
 |
| Fig.26.2 |
■ That means, we subtract the 'required term from the last' from the total and then add 1
• Similarly, in fig.b, If we want the 9th term from the last, it will be the (12-9+1)tth term = 4th term from the first.
9. So we want the 15th term.
It is given by: nth term = a + (n-1)d ⇒ 15th term = 10 + (15-1) ×-3 = 10 + 14 × -3 = 10 - 42 = -32Solved example 26.12
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Solution:
1. Let us write all the given values in the form of a series:
23, 21, 19, . . . ,5
2. This is an AP with d = -2. Because every term can be obtained by adding (-2) to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term 5
4. We have: nth term = a + (n-1)d ⇒ 5 = 23 + (n-1) ×-2
⇒ 5 = 23 - 2n + 2 ⇒ 2n = 23 - 5 + 2 ⇒ 2n = 20 ⇒ n = 10
• So there are a total of 10 rows in the flower bed.
Solved example 26.13
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
We have: nth term = an = a + (n-1)d
(i) a, d and n are given. an is the unknown. We can use the formula directly.
an = 7 + (8-1)×3 = 7 + 7 × 3 = 7 + 21 = 28
(ii) Here d is the unknown
0 = -18 + (10-1)×d ⇒ 0 = -18 + 9d ⇒ 9d = 18 ⇒ d = 2
(iii) Here a is the unknown
-5 = a + (18-1)×-3 ⇒ -5 = a - 51 ⇒ a = 51 -5 = 46
(iv) Here n is the unknown
3.6 = -18.9 + (n-1)×2.5 ⇒ 3.6 = -18.9 + 2.5n -2.5 ⇒ 3.6 = -21.4 + 2.5n ⇒ 2.5n = 25 ⇒ n = 10
(v) a, d and n are given. an is the unknown. We can use the formula directly.
an = 3.5 + (105-1)×0 = 3.5
• Also note that, since d is zero, all terms will be equal to the first term. So 105th term is also equal to 3.5
Solved example 26.14
Which term of the AP: 3, 8, 13, 18, . . . is 78
Solution:
1. 1st term a = 3, common difference d = 18 - 13 = 5
2. So we can write: nth term = a + (n-1)d ⇒ nth term = 3 + (n-1) ×5
3. Let 78 be the nth term. Then we can write:
78 = 3 + (n-1) ×5 ⇒ 78 = 3 + 5n - 5 ⇒ 5n = 78 + 5 - 3 ⇒ 5n = 80 ⇒ n = 16
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