In the previous section we completed a discussion on some topics in statistics. In this chapter we will learn about Arithmetic progressions. A general discussion about Sequences is given in Section 26.6 of this chapter.
Consider the following situation:
1. Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of Rs 8000, with an annual increment of Rs 500 in her salary.
2. From the above information, we take out two important points:
• Starting monthly salary = Rs 8000
• Annual increment = Rs 500
3. With the above two information, we can calculate the following:
• Monthly salary for the 1st year = Rs 8000
• Monthly salary for the 2nd year = 8000 + 500 = Rs 8500
• Monthly salary for the 3rd year = 8500 + 500 = Rs 9000
• Monthly salary for the 4th year = 9000 + 500 = Rs 9500
• Monthly salary for the 5th year = 9500 + 500 = Rs 10000
so on...
4. Now take out the final results in the above step (3) and write them in the form of a series. We get:
8000, 8500, 9000, 9500, . . . .
5. We can say: The terms in the above series are the salary obtained by Reena for the 1st, 2nd, 3rd, 4th, . . . years
Another example:
1. Consider the ladder in fig.26.1 below. The width of the ladder is decreasing as we go up.
This is because, the length of steps are decreasing uniformly by 2 cm as we go up. The bottom most step is 45 cm in length.
2. From the above information, we take out two important points:
• Length of the bottom most step = 45 cm
• Decrease in each successive step = 2 cm
3. With the above two information, we can calculate the following:
• Length of the bottom most step = 45 cm
• Length of the 2nd step = 45 – 2 = 43 cm
• Length of the 3rd step = 43 – 2 = 41 cm
• Length of the 4th step = 41 – 2 = 39 cm
so on . . .
4. Now take out the final results in the above step (3) and write them in the form of a series. We get:
45, 43, 41, 39, . . .
5. We can say: The terms in the above series are the lengths of 1st, 2nd, 3rd, 4th, . . . steps of the ladder
Another example
1. In a savings scheme, the amount increases by 25% after every three years. A person deposited Rs 8000.
2. From the above information, we take out two important points:
• Initial deposit = Rs 8000
• Percentage increase in every 3 years = 25%
3. With the above two information, we can calculate the following:
• Initial amount = Rs 8000
• Increase in amount after the 1st three years = 8000 × 25⁄100 = Rs 2000
• Total amount after the 1st 1st three years = 8000 + 2000 = 10000
[The last two steps above can be combined into a single step:
Final amount = initial amount + increase = (8000 + 8000 × 25⁄100) = 8000 (1 + 25⁄100)
= 8000 (1 + 1⁄4) = 8000 × 5⁄4 = 10000]
• Increase in amount after the 2nd three years (that is., after 6 years from the beginning) = 10000 × 25⁄100 = Rs 2500
• Total amount after the 2nd three years = 10000 + 2500 = 12500
[The above two steps can be combined into a single step:
Final amount = initial amount + increase = (10000 + 10000 × 25⁄100) = 10000 (1 + 25⁄100)
= 10000 (1 + 1⁄4) = 10000 × 5⁄4 = 12500]
• Increase in amount after the 3rd three years (that is., after 9 years from the beginning) = 12500 × 25⁄100 = Rs 3125
• Total amount after the 3rd three years = 12500 + 3125 = 15625
[The above two steps can be combined into a single step:
Final amount = initial amount + increase = (12500 + 12500 × 25⁄100) = 12500 (1 + 25⁄100)
= 12500 (1 + 1⁄4) = 12500 × 5⁄4 = 15625]
• Increase in amount after the 4th three years (that is., after 12 years from the beginning) = 15625 × 25⁄100 = Rs 3906.25
• Total amount after the 3rd three years = 15625 + 3906.25 = 19531.25
[The above two steps can be combined into a single step:
Final amount = initial amount + increase = (15625 + 15625 ×25⁄100) = 15625 × 5⁄4 = 19531.25]
4. Now take out the final results in the above step (3) and write them in the form of a series. We get:
8000, 10000, 12500, 15625, 19531.25, . . .
5. We can say: The terms in the above series are the initial amount and the amounts obtained at the end of 3, 6, 9, 12, . . .years
We come across such series in our day to day life. So we must learn more details about them.
■ Consider the first example:
• We have a first term 8000
• We obtained the successive terms by adding a fixed number 500 to the previous term
■ Consider the second example:
• We have a first term 45
• We obtained the successive terms by subtracting 2 from the previous term
• Subtracting 2 is same as adding (-2). So we can write:
• We obtained the successive terms by adding a fixed number (-2) to the previous term
■ Consider the third example:
• We have a first term 8000
• We obtained the successive terms by multiplying the previous term by a fixed number 5⁄4
■ An Arithmetic Progression is a series in which each term (except the first term) is obtained by adding a fixed number to the preceding term.
• The fixed number is called the common difference.
♦ The common difference can be positive, negative or zero.
• Let us analyse the series:
The 1st term is 1
The 2nd term is 2. The difference between the 2nd and 1st terms is 2-1 = 1
The 3rd term is 3. The difference between the 3rd and 2nd terms is 3-2 = 1
The 4th term is 4. The difference between the 4th and 3rd terms is 4-3 = 1
• So we see that successive terms are obtained by adding a fixed number '1' to the preceding terms.
• The given series is an AP. It's common difference is 1
(ii) 100, 70, 40, 10, . . .
• Let us analyse the series:
The 1st term is 100
The 2nd term is 70. The difference between the 2nd and 1st terms is 70-100 = -30
The 3rd term is 40. The difference between the 3rd and 2nd terms is 40-70 = -30
The 4th term is 10. The difference between the 4th and 3rd terms is 10-40 = -30
• So we see that successive terms are obtained by adding a fixed number '-30' to the preceding terms.
• The given series is an AP. It's common difference is -30
(iii) -3, -2, -1, 0, . . .
• Let us analyse the series:
The 1st term is -3
The 2nd term is -2. The difference between the 2nd and 1st terms is -2-(-3) = -2+3 = 1
The 3rd term is -1. The difference between the 3rd and 2nd terms is -1-(-2) = -1+2 = 1
The 4th term is 0. The difference between the 4th and 3rd terms is 0-(-1) = 0+1 = 1
• So we see that successive terms are obtained by adding a fixed number '1' to the preceding terms.
• The given series is an AP. It's common difference is 1
(iv) 3, 3, 3, 3, . . .
• Let us analyse the series:
The 1st term is 3
The 2nd term is 3. The difference between the 2nd and 1st terms is 3-3 = 0
The 3rd term is 3. The difference between the 3rd and 2nd terms is 3-3 = 0
The 4th term is 3. The difference between the 4th and 3rd terms is 3-3 = 0
• So we see that successive terms are obtained by adding a fixed number '0' to the preceding terms.
• The given series is an AP. It's common difference is 0
(v) -1, -1.5, -2, -2.5, . . .
• Let us analyse the series:
The 1st term is -1
The 2nd term is -1.5. The difference between the 2nd and 1st terms is -1.5-(-1) = -1.5+1 = -0.5
The 3rd term is -2. The difference between the 3rd and 2nd terms is -2-(-1.5) = -2+1.5 = -0.5
The 4th term is -2.5. The difference between the 4th and 3rd terms is -2.5-(-2) = -2.5+2 = -0.5
• So we see that successive terms are obtained by adding a fixed number '-0.5' to the preceding terms.
• The given series is an AP. It's common difference is -0.5
• It is an AP. Now, The carpenter wants to know the total length of all the steps. So that he can buy that much length of timber.
• For getting the total length, we must get the sum of all the terms in the AP.
• When we learn more details about AP, we will see easy methods to get such a sum.
■ Another situation that may arise:
• The carpenter knows that the width of the ladder goes on decreasing.
• Will it decrease to so small a value that, the ladder becomes smaller than a person climbing it?
• To know that, the carpenter should know the length of the top most step. This will be the least width of the ladder.
• That is., he must know the last term of the series. He must know it before starting to make the ladder.
• When we learn more details about AP, we will see easy methods to get the last term. In fact, we will be able to get any term we want.
• Let the 3rd term be a3
So on . . .
• Let the nth term be an
■ Then the AP becomes: a1, a2, a3, . . . , an.
■ We will also get:
• a2 - a1 = d
• a3 - a2 = d
• a4 - a3 = d
so on . . .
• an - an-1 = d
■ Now we get an interesting result:
• Consider a2. From the above steps we get: a2 = a1 +d
• Consider a3. From the above steps we get: a3 = a2 +d
♦ But a2 = a1 +d
♦ So a3 = (a1 +d) + d = a1 + 2d
• Consider a4. We have: a4 = a3 +d
♦ But a3 = a1 +2d
♦ So a4 = (a1 + 2d) + d = a1 + 3d
■ Based on the above, we can write the AP as: a, a+d, a+2d, a+3d, . . .
This is called the general form of an AP.
The common difference 'd' of the AP
If we know both of them, we can completely write that AP
Let us write some examples:
Example (i): Given a = 6 and d = 3. Write the AP
Solution:
a1 = a = 6
a2 = a1 + d = 6 + 3 = 9
a3 = a1 + 2d = 6 + 2 × 3 = 12
a4 = a1 + 3d = 6 + 3 × 3 = 15
So on . . .
Thus the required AP is: 6, 9, 12, 15, . . .
Example (ii): Given a = 1 and d = 0.1. Write the AP
Solution:
a1 = a = 1
a2 = a1 + d = 1 + 0.1 = 1.1
a3 = a1 + 2d = 1 + 2 × 0.1 = 1.2
a4 = a1 + 3d = 1 + 3 × 0.1 = 1.3
So on . . .
Thus the required AP is: 1, 1.1, 1.2, 1.3, . . .
So we have learned to write an AP when a and d are given. We must be able to work in the reverse order also. We will see it in the next section.
Consider the following situation:
1. Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of Rs 8000, with an annual increment of Rs 500 in her salary.
2. From the above information, we take out two important points:
• Starting monthly salary = Rs 8000
• Annual increment = Rs 500
3. With the above two information, we can calculate the following:
• Monthly salary for the 1st year = Rs 8000
• Monthly salary for the 2nd year = 8000 + 500 = Rs 8500
• Monthly salary for the 3rd year = 8500 + 500 = Rs 9000
• Monthly salary for the 4th year = 9000 + 500 = Rs 9500
• Monthly salary for the 5th year = 9500 + 500 = Rs 10000
so on...
4. Now take out the final results in the above step (3) and write them in the form of a series. We get:
8000, 8500, 9000, 9500, . . . .
5. We can say: The terms in the above series are the salary obtained by Reena for the 1st, 2nd, 3rd, 4th, . . . years
Another example:
1. Consider the ladder in fig.26.1 below. The width of the ladder is decreasing as we go up.
 |
| Fig.26.1 |
2. From the above information, we take out two important points:
• Length of the bottom most step = 45 cm
• Decrease in each successive step = 2 cm
3. With the above two information, we can calculate the following:
• Length of the bottom most step = 45 cm
• Length of the 2nd step = 45 – 2 = 43 cm
• Length of the 3rd step = 43 – 2 = 41 cm
• Length of the 4th step = 41 – 2 = 39 cm
so on . . .
4. Now take out the final results in the above step (3) and write them in the form of a series. We get:
45, 43, 41, 39, . . .
5. We can say: The terms in the above series are the lengths of 1st, 2nd, 3rd, 4th, . . . steps of the ladder
Another example
1. In a savings scheme, the amount increases by 25% after every three years. A person deposited Rs 8000.
2. From the above information, we take out two important points:
• Initial deposit = Rs 8000
• Percentage increase in every 3 years = 25%
3. With the above two information, we can calculate the following:
• Initial amount = Rs 8000
• Increase in amount after the 1st three years = 8000 × 25⁄100 = Rs 2000
• Total amount after the 1st 1st three years = 8000 + 2000 = 10000
[The last two steps above can be combined into a single step:
Final amount = initial amount + increase = (8000 + 8000 × 25⁄100) = 8000 (1 + 25⁄100)
= 8000 (1 + 1⁄4) = 8000 × 5⁄4 = 10000]
• Increase in amount after the 2nd three years (that is., after 6 years from the beginning) = 10000 × 25⁄100 = Rs 2500
• Total amount after the 2nd three years = 10000 + 2500 = 12500
[The above two steps can be combined into a single step:
Final amount = initial amount + increase = (10000 + 10000 × 25⁄100) = 10000 (1 + 25⁄100)
= 10000 (1 + 1⁄4) = 10000 × 5⁄4 = 12500]
• Increase in amount after the 3rd three years (that is., after 9 years from the beginning) = 12500 × 25⁄100 = Rs 3125
• Total amount after the 3rd three years = 12500 + 3125 = 15625
[The above two steps can be combined into a single step:
Final amount = initial amount + increase = (12500 + 12500 × 25⁄100) = 12500 (1 + 25⁄100)
= 12500 (1 + 1⁄4) = 12500 × 5⁄4 = 15625]
• Increase in amount after the 4th three years (that is., after 12 years from the beginning) = 15625 × 25⁄100 = Rs 3906.25
• Total amount after the 3rd three years = 15625 + 3906.25 = 19531.25
[The above two steps can be combined into a single step:
Final amount = initial amount + increase = (15625 + 15625 ×25⁄100) = 15625 × 5⁄4 = 19531.25]
4. Now take out the final results in the above step (3) and write them in the form of a series. We get:
8000, 10000, 12500, 15625, 19531.25, . . .
5. We can say: The terms in the above series are the initial amount and the amounts obtained at the end of 3, 6, 9, 12, . . .years
■ Consider the first example:
• We have a first term 8000
• We obtained the successive terms by adding a fixed number 500 to the previous term
■ Consider the second example:
• We have a first term 45
• We obtained the successive terms by subtracting 2 from the previous term
• Subtracting 2 is same as adding (-2). So we can write:
• We obtained the successive terms by adding a fixed number (-2) to the previous term
■ Consider the third example:
• We have a first term 8000
• We obtained the successive terms by multiplying the previous term by a fixed number 5⁄4
■ In this chapter we will be learning only those series in which successive terms are obtained by adding a fixed number. There is a special name for such series: Arithmetic progressions. In short form it is written as AP
Let us write it’s definition:■ An Arithmetic Progression is a series in which each term (except the first term) is obtained by adding a fixed number to the preceding term.
• The fixed number is called the common difference.
♦ The common difference can be positive, negative or zero.
Let us see a few more examples:
(i) 1, 2, 3, 4, . . .• Let us analyse the series:
The 1st term is 1
The 2nd term is 2. The difference between the 2nd and 1st terms is 2-1 = 1
The 3rd term is 3. The difference between the 3rd and 2nd terms is 3-2 = 1
The 4th term is 4. The difference between the 4th and 3rd terms is 4-3 = 1
• So we see that successive terms are obtained by adding a fixed number '1' to the preceding terms.
• The given series is an AP. It's common difference is 1
(ii) 100, 70, 40, 10, . . .
• Let us analyse the series:
The 1st term is 100
The 2nd term is 70. The difference between the 2nd and 1st terms is 70-100 = -30
The 3rd term is 40. The difference between the 3rd and 2nd terms is 40-70 = -30
The 4th term is 10. The difference between the 4th and 3rd terms is 10-40 = -30
• So we see that successive terms are obtained by adding a fixed number '-30' to the preceding terms.
• The given series is an AP. It's common difference is -30
(iii) -3, -2, -1, 0, . . .
• Let us analyse the series:
The 1st term is -3
The 2nd term is -2. The difference between the 2nd and 1st terms is -2-(-3) = -2+3 = 1
The 3rd term is -1. The difference between the 3rd and 2nd terms is -1-(-2) = -1+2 = 1
The 4th term is 0. The difference between the 4th and 3rd terms is 0-(-1) = 0+1 = 1
• So we see that successive terms are obtained by adding a fixed number '1' to the preceding terms.
• The given series is an AP. It's common difference is 1
(iv) 3, 3, 3, 3, . . .
• Let us analyse the series:
The 1st term is 3
The 2nd term is 3. The difference between the 2nd and 1st terms is 3-3 = 0
The 3rd term is 3. The difference between the 3rd and 2nd terms is 3-3 = 0
The 4th term is 3. The difference between the 4th and 3rd terms is 3-3 = 0
• So we see that successive terms are obtained by adding a fixed number '0' to the preceding terms.
• The given series is an AP. It's common difference is 0
(v) -1, -1.5, -2, -2.5, . . .
• Let us analyse the series:
The 1st term is -1
The 2nd term is -1.5. The difference between the 2nd and 1st terms is -1.5-(-1) = -1.5+1 = -0.5
The 3rd term is -2. The difference between the 3rd and 2nd terms is -2-(-1.5) = -2+1.5 = -0.5
The 4th term is -2.5. The difference between the 4th and 3rd terms is -2.5-(-2) = -2.5+2 = -0.5
• So we see that successive terms are obtained by adding a fixed number '-0.5' to the preceding terms.
• The given series is an AP. It's common difference is -0.5
■ Consider the example of the ladder that we saw above. We obtained a series:
45, 43, 41, 39, . . .• It is an AP. Now, The carpenter wants to know the total length of all the steps. So that he can buy that much length of timber.
• For getting the total length, we must get the sum of all the terms in the AP.
• When we learn more details about AP, we will see easy methods to get such a sum.
■ Another situation that may arise:
• The carpenter knows that the width of the ladder goes on decreasing.
• Will it decrease to so small a value that, the ladder becomes smaller than a person climbing it?
• To know that, the carpenter should know the length of the top most step. This will be the least width of the ladder.
• That is., he must know the last term of the series. He must know it before starting to make the ladder.
• When we learn more details about AP, we will see easy methods to get the last term. In fact, we will be able to get any term we want.
• Let the 1st term of an AP be a1
• Let the 2nd term be a2• Let the 3rd term be a3
So on . . .
• Let the nth term be an
■ Then the AP becomes: a1, a2, a3, . . . , an.
■ We will also get:
• a2 - a1 = d
• a3 - a2 = d
• a4 - a3 = d
so on . . .
• an - an-1 = d
■ Now we get an interesting result:
• Consider a2. From the above steps we get: a2 = a1 +d
• Consider a3. From the above steps we get: a3 = a2 +d
♦ But a2 = a1 +d
♦ So a3 = (a1 +d) + d = a1 + 2d
• Consider a4. We have: a4 = a3 +d
♦ But a3 = a1 +2d
♦ So a4 = (a1 + 2d) + d = a1 + 3d
■ Based on the above, we can write the AP as: a, a+d, a+2d, a+3d, . . .
This is called the general form of an AP.
To write an AP, we will need both the following two information:
The first term 'a' of the APThe common difference 'd' of the AP
If we know both of them, we can completely write that AP
Let us write some examples:
Example (i): Given a = 6 and d = 3. Write the AP
Solution:
a1 = a = 6
a2 = a1 + d = 6 + 3 = 9
a3 = a1 + 2d = 6 + 2 × 3 = 12
a4 = a1 + 3d = 6 + 3 × 3 = 15
So on . . .
Thus the required AP is: 6, 9, 12, 15, . . .
Example (ii): Given a = 1 and d = 0.1. Write the AP
Solution:
a1 = a = 1
a2 = a1 + d = 1 + 0.1 = 1.1
a3 = a1 + 2d = 1 + 2 × 0.1 = 1.2
a4 = a1 + 3d = 1 + 3 × 0.1 = 1.3
So on . . .
Thus the required AP is: 1, 1.1, 1.2, 1.3, . . .
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