Hello students, welcome to today's mathematics lesson. I am so happy to be here with you to learn about one of the most interesting and useful topics in algebra - Arithmetic Progressions. You will see how this concept appears in many real-life situations, and by the end of this chapter, you will be able to solve problems that seem complicated at first glance but become quite straightforward once you understand the patterns.
Let us start by thinking about the world around us. You must have observed that in nature, many things follow a certain pattern. Look at the petals of a sunflower - they are arranged in a beautiful spiral pattern. Notice the holes of a honeycomb - they are perfectly hexagonal and arranged in a pattern. Look at the grains on a maize cob, the spirals on a pineapple, and on a pine cone - all these follow mathematical patterns. Isn't it wonderful how mathematics is hidden in nature?
Now, let us look for some patterns which occur in our day-to-day life. I want you to think about these examples carefully.
First example: Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of ₹8000, with an annual increment of ₹500 in her salary. What will be her salary for the first year, second year, third year, and so on? Let us think. In the first year, her salary is ₹8000. In the second year, it becomes ₹8000 plus ₹500, which is ₹8500. In the third year, it becomes ₹8500 plus ₹500, which is ₹9000. So the salaries are ₹8000, ₹8500, ₹9000, and this pattern continues. Do you see how each term is obtained? We are adding ₹500 each time to get the next term.
Second example: The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top. The bottom rung is 45 cm in length. What are the lengths of the rungs? The first rung from the bottom is 45 cm. The second rung is 2 cm less, so 43 cm. The third rung is again 2 cm less, so 41 cm. Continuing this pattern, we get 45, 43, 41, 39, 37, 35, 33, 31. Here also, we are adding a fixed number - but in this case, we are adding negative 2, which means we are subtracting 2 each time.
Third example: Shakila puts ₹100 into her daughter's money box when she was one year old and increased the amount by ₹50 every year. So on her first birthday, she puts ₹100, on the second birthday ₹150, on the third birthday ₹200, on the fourth birthday ₹250, and so on. Again, we are adding ₹50 each time.
Fourth example: In a savings scheme, the amount becomes 5/4 times of itself after every 3 years. The maturity amount of an investment of ₹8000 after 3, 6, 9, and 12 years will be ₹10000, ₹12500, ₹15625, and ₹19531.25. Here, each term is obtained by multiplying the previous term by 5/4, not by adding a fixed number.
Fifth example: The number of unit squares in squares with side 1, 2, 3, ... units are 1², 2², 3², ... which are 1, 4, 9, 16, and so on. Here, each term is the square of consecutive numbers.
Sixth example: A pair of rabbits are too young to produce in their first month. In the second month, and every subsequent month, they produce a new pair. Each new pair of rabbits produce a new pair in their second month and in every subsequent month. Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd, ... 6th month are 1, 1, 2, 3, 5, 8. This is the famous Fibonacci sequence!
Now, students, look at these examples carefully. In some, we find that the succeeding terms are obtained by adding a fixed number - like in Reena's salary, the ladder rungs, and Shakila's money box. In others, we find they are obtained by multiplying with a fixed number, or they are squares of consecutive numbers, and so on.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems. This pattern is called an Arithmetic Progression.
Now, let us understand what exactly is an Arithmetic Progression. Consider the following lists of numbers:
First list: 1, 2, 3, 4, ... Second list: 100, 70, 40, 10, ... Third list: -3, -2, -1, 0, ... Fourth list: 3, 3, 3, 3, ... Fifth list: -1.0, -1.5, -2.0, -2.5, ...
Each of the numbers in these lists is called a term. Now, given a term, can you write the next term in each of the lists? If so, how will you write it? We follow a pattern or rule.
In the first list, each term is 1 more than the term preceding it. So from 1 we get 2, from 2 we get 3, and so on.
In the second list, each term is 30 less than the term preceding it. So from 100 we get 70, from 70 we get 40.
In the third list, each term is obtained by adding 1 to the term preceding it. So from -3 we get -2, from -2 we get -1.
In the fourth list, all the terms are 3. This means each term is obtained by adding or subtracting 0 to the term preceding it. So the difference is 0.
In the fifth list, each term is obtained by adding -0.5, that is, subtracting 0.5 from the term preceding it. So from -1.0 we get -1.5, from -1.5 we get -2.0.
In all these lists, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such a list of numbers is said to form an Arithmetic Progression, which we often write as AP.
So, students, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term. This fixed number is called the common difference of the AP. Remember that it can be positive, negative, or zero. If the common difference is positive, the terms increase. If it is negative, the terms decrease. If it is zero, all terms are equal.
Now, let us denote the first term of an AP by a₁, the second term by a₂, the nth term by aₙ, and the common difference by d. Then the AP becomes a₁, a₂, a₃, ..., aₙ.
So, a₂ minus a₁ equals a₃ minus a₂, and so on, equals aₙ minus aₙ₋₁, and all of these equal d. This is the defining property of an arithmetic progression.
Let me give you some more examples of AP. The heights in cm of some students of a school standing in a queue in the morning assembly are 147, 148, 149, ..., 157. Here, each height increases by 1 cm, so it is an AP with first term 147 and common difference 1.
The minimum temperatures recorded for a week in the month of January in a city, arranged in ascending order, are -3.1, -3.0, -2.9, -2.8, -2.7, -2.6, -2.5. Here, each temperature increases by 0.1 degrees, so it is an AP with first term -3.1 and common difference 0.1.
The balance money after paying 5% of the total loan of ₹1000 every month is 950, 900, 850, 800, ..., 50. Here, each month the balance decreases by ₹50, so it is an AP with first term 950 and common difference -50.
The cash prizes given by a school to the toppers of Classes I to XII are respectively ₹200, ₹250, ₹300, ₹350, ..., ₹750. Here, each prize increases by ₹50, so it is an AP with first term 200 and common difference 50.
The total savings after every month for 10 months when ₹50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500. Here also, each month the savings increase by ₹50, so it is an AP with first term 50 and common difference 50.
Now, students, you might be wondering - can we represent an AP in a general form? Yes, we can! The general form of an AP is a, a + d, a + 2d, a + 3d, ... where a is the first term and d is the common difference. This is called the general form of an AP.
Now, let me ask you a question. To know about an AP, what is the minimum information that you need? Is it enough to know the first term? Or is it enough to know only the common difference? You will find that you need to know both - the first term a and the common difference d.
For instance, if the first term a is 6 and the common difference d is 3, then the AP is 6, 9, 12, 15, ... And if a is 6 and d is -3, then the AP is 6, 3, 0, -3, ... Similarly, when a = -7, d = -2, the AP is -7, -9, -11, -13, ... When a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3, ... When a = 0, d = 1½, the AP is 0, 1½, 3, 4½, 6, ... And when a = 2, d = 0, the AP is 2, 2, 2, 2, ...
So, if you know what a and d are, you can list the AP. What about the other way round? That is, if you are given a list of numbers, can you say that it is an AP and then find a and d? Since a is the first term, it can easily be written. We know that in an AP, every succeeding term is obtained by adding d to the preceding term. So, d can be found by subtracting any term from its succeeding term, that is, the term which immediately follows it. The difference should be the same for an AP.
For example, for the list of numbers 6, 9, 12, 15, ..., we have a₂ minus a₁ equals 9 minus 6 equals 3, a₃ minus a₂ equals 12 minus 9 equals 3, a₄ minus a₃ equals 15 minus 12 equals 3. Here, the difference of any two consecutive terms is 3. So, the given list is an AP whose first term a is 6 and common difference d is 3.
For the list of numbers 6, 3, 0, -3, ..., we have a₂ minus a₁ equals 3 minus 6 equals -3, a₃ minus a₂ equals 0 minus 3 equals -3, a₄ minus a₃ equals -3 minus 0 equals -3. Similarly, this is also an AP whose first term is 6 and the common difference is -3.
In general, for an AP a₁, a₂, ..., aₙ, we have d equals aₖ₊₁ minus aₖ, where aₖ₊₁ and aₖ are the (k+1)th and the kth terms respectively.
To obtain d in a given AP, we need not find all of a₂ minus a₁, a₃ minus a₂, a₄ minus a₃, and so on. It is enough to find only one of them.
Consider the list of numbers 1, 1, 2, 3, 5, ... By looking at it, you can tell that the difference between any two consecutive terms is not the same. So, this is not an AP.
Note that to find d in the AP 6, 3, 0, -3, ..., we have subtracted 6 from 3 and not 3 from 6. That is, we should subtract the kth term from the (k+1)th term, even if the (k+1)th term is smaller. This is important, students. The common difference is always calculated as the next term minus the current term, not the other way around.
Now, let us look at some examples from your textbook to make this concept clearer.
Example 1: For the AP: 3/2, 1/2, -1/2, -3/2, ..., write the first term a and the common difference d.
Here, a = 3/2. To find d, we take any two consecutive terms. Let us take 1/2 minus 3/2, which equals -1. So d = -1. Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP.
Example 2: Which of the following list of numbers form an AP? If they form an AP, write the next two terms.
First list: 4, 10, 16, 22, ... We have a₂ minus a₁ equals 10 minus 4 equals 6, a₃ minus a₂ equals 16 minus 10 equals 6, a₄ minus a₃ equals 22 minus 16 equals 6. So aₖ₊₁ minus aₖ is the same every time. So, the given list of numbers forms an AP with the common difference d = 6. The next two terms are: 22 plus 6 equals 28, and 28 plus 6 equals 34.
Second list: 1, -1, -3, -5, ... We have a₂ minus a₁ equals -1 minus 1 equals -2, a₃ minus a₂ equals -3 minus (-1) equals -3 plus 1 equals -2, a₄ minus a₃ equals -5 minus (-3) equals -5 plus 3 equals -2. So aₖ₊₁ minus aₖ is the same every time. So, the given list of numbers forms an AP with the common difference d = -2. The next two terms are: -5 plus (-2) equals -7, and -7 plus (-2) equals -9.
Third list: -2, 2, -2, 2, -2, ... We have a₂ minus a₁ equals 2 minus (-2) equals 2 plus 2 equals 4, but a₃ minus a₂ equals -2 minus 2 equals -4. As a₂ minus a₁ is not equal to a₃ minus a₂, the given list of numbers does not form an AP.
Fourth list: 1, 1, 1, 2, 2, 2, 3, 3, 3, ... We have a₂ minus a₁ equals 1 minus 1 equals 0, a₃ minus a₂ equals 1 minus 1 equals 0, but a₄ minus a₃ equals 2 minus 1 equals 1. Here, a₂ minus a₁ equals a₃ minus a₂, but it is not equal to a₄ minus a₃. So, the given list of numbers does not form an AP.
So, students, remember that for a list to be an AP, the difference between consecutive terms must be the same throughout the entire list.
Now, let me introduce you to another important concept - the concept of finite and infinite APs. In some of our examples, there are only a finite number of terms. For instance, the heights of students from 147 cm to 157 cm, or the cash prizes from Class I to Class XII. Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions has a last term.
On the other hand, the APs like 1, 2, 3, 4, ... or 100, 70, 40, 10, ... or -3, -2, -1, 0, ... are not finite APs. They go on forever. These are called infinite Arithmetic Progressions. Such APs do not have a last term.
Now, let us move on to a very important concept - the nth term of an AP. This is also called the general term of the AP.
Let us consider the situation again in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of ₹8000, with an annual increment of ₹500. What would be her monthly salary for the fifth year?
To answer this, let us first see what her monthly salary for the second year would be. It would be ₹8000 plus ₹500, which is ₹8500. In the same way, we can find the monthly salary for the 3rd, 4th, and 5th year by adding ₹500 to the salary of the previous year.
So, the salary for the third year equals ₹8500 plus ₹500, which equals ₹9000. We can also write this as ₹8000 plus 2 times ₹500, which is ₹8000 plus (3 minus 1) times ₹500.
The salary for the fourth year equals ₹9000 plus ₹500, which equals ₹8000 plus 3 times ₹500, which is ₹8000 plus (4 minus 1) times ₹500.
The salary for the fifth year equals ₹9500 plus ₹500, which equals ₹8000 plus 4 times ₹500, which is ₹8000 plus (5 minus 1) times ₹500.
So, we can see a pattern emerging. The salary for the nth year would be ₹8000 plus (n minus 1) times ₹500.
Now, let us generalize this. Let a₁, a₂, a₃, ... be an AP whose first term a₁ is a and the common difference is d.
Then, the second term a₂ equals a plus d, which we can write as a plus (2 minus 1) times d.
The third term a₃ equals a₂ plus d, which equals (a plus d) plus d, which equals a plus 2d, which we can write as a plus (3 minus 1) times d.
The fourth term a₄ equals a₃ plus d, which equals (a plus 2d) plus d, which equals a plus 3d, which we can write as a plus (4 minus 1) times d.
And so on.
Looking at the pattern, we can say that the nth term aₙ equals a plus (n minus 1) times d.
So, students, the nth term aₙ of the AP with first term a and common difference d is given by aₙ equals a plus (n minus 1) times d. This is one of the most important formulas in this chapter. Please remember it.
aₙ is also called the general term of the AP. If there are m terms in the AP, then aₘ represents the last term, which is sometimes also denoted by l.
Now, let us look at some examples to understand how to use this formula.
Example 3: Find the 10th term of the AP: 2, 7, 12, ...
Here, a equals 2, d equals 7 minus 2 equals 5, and n equals 10.
We have aₙ equals a plus (n minus 1) times d.
So, a₁₀ equals 2 plus (10 minus 1) times 5, which equals 2 plus 45, which equals 47.
Therefore, the 10th term of the given AP is 47.
Example 4: Which term of the AP: 21, 18, 15, ... is -81? Also, is any term 0? Give reason for your answer.
Here, a equals 21, d equals 18 minus 21 equals -3, and aₙ equals -81. We have to find n.
As aₙ equals a plus (n minus 1) times d, we have -81 equals 21 plus (n minus 1) times (-3).
So, -81 equals 24 minus 3n.
So, -105 equals -3n.
Therefore, n equals 35.
So, the 35th term of the given AP is -81.
Next, we want to know if there is any n for which aₙ equals 0. If such an n is there, then 21 plus (n minus 1) times (-3) equals 0.
So, 3 times (n minus 1) equals 21.
So, n minus 1 equals 7.
So, n equals 8.
Therefore, the eighth term is 0.
Example 5: Determine the AP whose 3rd term is 5 and the 7th term is 9.
We have a₃ equals a plus (3 minus 1) times d, which is a plus 2d, and this equals 5. This is our first equation.
We also have a₇ equals a plus (7 minus 1) times d, which is a plus 6d, and this equals 9. This is our second equation.
Solving these two equations, we get a equals 3 and d equals 1.
Hence, the required AP is 3, 4, 5, 6, 7, ...
Example 6: Check whether 301 is a term of the list of numbers 5, 11, 17, 23, ...
We have a₂ minus a₁ equals 11 minus 5 equals 6, a₃ minus a₂ equals 17 minus 11 equals 6, a₄ minus a₃ equals 23 minus 17 equals 6. As aₖ₊₁ minus aₖ is the same for k equals 1, 2, 3, and so on, the given list of numbers is an AP.
Now, a equals 5 and d equals 6.
Let 301 be a term, say, the nth term of this AP.
We know that aₙ equals a plus (n minus 1) times d.
So, 301 equals 5 plus (n minus 1) times 6.
That is, 301 equals 6n minus 1.
So, n equals 302 divided by 6, which equals 151 divided by 3.
But n should be a positive integer. Since 151 divided by 3 is not an integer, 301 is not a term of the given list of numbers.
Example 7: How many two-digit numbers are divisible by 3?
The list of two-digit numbers divisible by 3 is: 12, 15, 18, ..., 99.
Is this an AP? Yes, it is. Here, a equals 12, d equals 3, and aₙ equals 99.
As aₙ equals a plus (n minus 1) times d, we have 99 equals 12 plus (n minus 1) times 3.
That is, 87 equals (n minus 1) times 3.
So, n minus 1 equals 87 divided by 3, which equals 29.
So, n equals 30.
Therefore, there are 30 two-digit numbers divisible by 3.
Example 8: Find the 11th term from the last term (towards the first term) of the AP: 10, 7, 4, ..., -62.
Here, a equals 10, d equals 7 minus 10 equals -3, and l equals -62, where l equals a plus (n minus 1) times d.
To find the 11th term from the last term, we will first find the total number of terms in the AP.
So, -62 equals 10 plus (n minus 1) times (-3).
That is, -72 equals (n minus 1) times (-3).
So, n minus 1 equals 24.
Or n equals 25.
So, there are 25 terms in the given AP.
Now, the 11th term from the last term will be the (25 minus 11 plus 1)th term, which is the 15th term. Note that it will not be the 14th term. Let me explain why. If we count from the last term, the last term itself is the 1st term from the last. So the 11th term from the last would be the (n minus 10)th term from the beginning. Since n equals 25, this is the 15th term.
So, a₁₅ equals 10 plus (15 minus 1) times (-3), which equals 10 minus 42, which equals -32.
So, the 11th term from the last term is -32.
There is also an alternative solution. If we write the given AP in the reverse order, then a becomes -62 and d becomes 3. So the question now becomes finding the 11th term with these a and d.
So, a₁₁ equals -62 plus (11 minus 1) times 3, which equals -62 plus 30, which equals -32.
So, the 11th term, which is now the required term, is -32.
Example 9: A sum of ₹1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
We know that the formula to calculate simple interest is given by Simple Interest equals (P times R times T) divided by 100.
So, the interest at the end of the 1st year equals ₹(1000 times 8 times 1) divided by 100, which is ₹80.
The interest at the end of the 2nd year equals ₹(1000 times 8 times 2) divided by 100, which is ₹160.
The interest at the end of the 3rd year equals ₹(1000 times 8 times 3) divided by 100, which is ₹240.
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
So, the interest in rupees at the end of the 1st, 2nd, 3rd, ... years, respectively, are 80, 160, 240, ...
It is an AP as the difference between the consecutive terms in the list is 80, that is, d equals 80. Also, a equals 80.
So, to find the interest at the end of 30 years, we shall find a₃₀.
Now, a₃₀ equals a plus (30 minus 1) times d, which equals 80 plus 29 times 80, which equals 2400.
So, the interest at the end of 30 years will be ₹2400.
Example 10: 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?
The number of rose plants in the 1st, 2nd, 3rd, ... rows are: 23, 21, 19, ..., 5.
It forms an AP. Let the number of rows in the flower bed be n.
Then a equals 23, d equals 21 minus 23 equals -2, and aₙ equals 5.
As aₙ equals a plus (n minus 1) times d, we have 5 equals 23 plus (n minus 1) times (-2).
That is, -18 equals (n minus 1) times (-2).
So, n equals 10.
Therefore, there are 10 rows in the flower bed.
Now, students, let me recap what we have learned so far. We learned that an arithmetic progression is a list of numbers where each term is obtained by adding a fixed number (called the common difference) to the previous term. We learned how to identify whether a given list is an AP, and how to find the first term and common difference. We also learned the very important formula for the nth term: aₙ equals a plus (n minus 1) times d. We saw several examples of how to use this formula to find any term of an AP, to find which term is a particular number, and to solve real-life problems.
Now, let us move on to another very important concept - the sum of the first n terms of an AP.
Let us consider the situation again in which Shakila put ₹100 into her daughter's money box when she was one year old, ₹150 on her second birthday, ₹200 on her third birthday, and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old?
Here, the amount of money in rupees put in the money box on her first, second, third, fourth, ... birthday were respectively 100, 150, 200, 250, ... till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don't you think it would be a tedious and time-consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see.
We consider the problem given to Gauss, to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how he did it? He wrote:
S equals 1 plus 2 plus 3 plus ... plus 99 plus 100.
And then, he reversed the numbers to write:
S equals 100 plus 99 plus ... plus 3 plus 2 plus 1.
Adding these two, he got:
2S equals (100 plus 1) plus (99 plus 2) plus ... plus (3 plus 98) plus (2 plus 99) plus (1 plus 100), which equals 101 plus 101 plus ... plus 101 plus 101, which is 101 added 100 times.
So, S equals 100 times 101 divided by 2, which equals 5050.
This is a brilliant technique! We will now use the same technique to find the sum of the first n terms of an AP: a, a plus d, a plus 2d, ...
The nth term of this AP is a plus (n minus 1) times d. Let S denote the sum of the first n terms of the AP. We have:
S equals a plus (a plus d) plus (a plus 2d) plus ... plus [a plus (n minus 1) times d]. This is equation (1).
Rewriting the terms in reverse order, we have:
S equals [a plus (n minus 1) times d] plus [a plus (n minus 2) times d] plus ... plus (a plus d) plus a. This is equation (2).
On adding (1) and (2), term-wise, we get:
2S equals [2a plus (n minus 1) times d] plus [2a plus (n minus 1) times d] plus ... plus [2a plus (n minus 1) times d] plus [2a plus (n minus 1) times d], which is (2a plus (n minus 1) times d) added n times.
Or, 2S equals n times [2a plus (n minus 1) times d].
Or, S equals n divided by 2 times [2a plus (n minus 1) times d].
So, the sum of the first n terms of an AP is given by:
S equals n/2 times [2a plus (n minus 1) times d].
We can also write this as S equals n/2 times [a plus a plus (n minus 1) times d], which is S equals n/2 times (a plus aₙ).
Now, if there are only n terms in an AP, then aₙ equals l, the last term. From this, we see that S equals n/2 times (a plus l). This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
Now we return to the question that was posed to us in the beginning. The amount of money in rupees in the money box of Shakila's daughter on 1st, 2nd, 3rd, 4th birthday, ... were 100, 150, 200, 250, ... respectively.
This is an AP. We have to find the total money collected on her 21st birthday, that is, the sum of the first 21 terms of this AP.
Here, a equals 100, d equals 50, and n equals 21. Using the formula S equals n/2 times [2a plus (n minus 1) times d], we have:
S equals 21/2 times [2 times 100 plus (21 minus 1) times 50], which equals 21/2 times [200 plus 1000], which equals 21/2 times 1200, which equals 12600.
So, the amount of money collected on her 21st birthday is ₹12600.
Hasn't the use of the formula made it much easier to solve the problem?
We also use Sₙ in place of S to denote the sum of first n terms of the AP. We write S₂₀ to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d, and n. If we know any three of them, we can find the fourth.
Remark: The nth term of an AP is the difference of the sum to first n terms and the sum to first (n minus 1) terms of it, that is, aₙ equals Sₙ minus Sₙ₋₁.
Now, let us look at some more examples to understand how to use the sum formula.
Example 11: Find the sum of the first 22 terms of the AP: 8, 3, -2, ...
Here, a equals 8, d equals 3 minus 8 equals -5, and n equals 22.
We know that S equals n/2 times [2a plus (n minus 1) times d].
Therefore, S equals 22/2 times [16 plus 21 times (-5)], which equals 11 times (16 minus 105), which equals 11 times (-89), which equals -979.
So, the sum of the first 22 terms of the AP is -979.
Example 12: If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Here, S₁₄ equals 1050, n equals 14, and a equals 10.
As Sₙ equals n/2 times [2a plus (n minus 1) times d], we have:
1050 equals 14/2 times [20 plus 13d], which equals 7 times (20 plus 13d), which equals 140 plus 91d.
So, 910 equals 91d.
Or, d equals 10.
Therefore, a₂₀ equals 10 plus (20 minus 1) times 10, which equals 10 plus 190, which equals 200. So the 20th term is 200.
Example 13: How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?
Here, a equals 24, d equals 21 minus 24 equals -3, and Sₙ equals 78. We need to find n.
We know that Sₙ equals n/2 times [2a plus (n minus 1) times d].
So, 78 equals n/2 times [48 plus (n minus 1) times (-3)], which equals n/2 times [51 minus 3n].
Or, 3n² minus 51n plus 156 equals 0.
Or, n² minus 17n plus 52 equals 0.
Or, (n minus 4) times (n minus 13) equals 0.
Or, n equals 4 or 13.
Both values of n are admissible. So, the number of terms is either 4 or 13.
Remarks: In this case, the sum of the first 4 terms equals the sum of the first 13 terms, which is 78. Two answers are possible because the sum of the terms from the 5th to the 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Example 14: Find the sum of the first 1000 positive integers. Also find the sum of the first n positive integers.
First part: Let S equals 1 plus 2 plus 3 plus ... plus 1000.
Using the formula Sₙ equals n/2 times (a plus l) for the sum of the first n terms of an AP, we have S₁₀₀₀ equals 1000/2 times (1 plus 1000), which equals 500 times 1001, which equals 500500.
So, the sum of the first 1000 positive integers is 500500.
Second part: Let Sₙ equals 1 plus 2 plus 3 plus ... plus n.
Here, a equals 1 and the last term l is n.
Therefore, Sₙ equals n times (1 plus n) divided by 2, or Sₙ equals n(n plus 1)/2.
So, the sum of first n positive integers is given by Sₙ equals n(n plus 1)/2.
Example 15: Find the sum of first 24 terms of the list of numbers whose nth term is given by aₙ equals 3 plus 2n.
As aₙ equals 3 plus 2n, we have a₁ equals 3 plus 2 equals 5, a₂ equals 3 plus 2 times 2 equals 7, a₃ equals 3 plus 2 times 3 equals 9, and so on.
The list of numbers becomes 5, 7, 9, 11, ...
Here, 7 minus 5 equals 9 minus 7 equals 11 minus 9 equals 2, and so on.
So, it forms an AP with common difference d equals 2.
To find S₂₄, we have n equals 24, a equals 5, d equals 2.
Therefore, S₂₄ equals 24/2 times [2 times 5 plus (24 minus 1) times 2], which equals 12 times [10 plus 46], which equals 12 times 56, which equals 672.
So, the sum of the first 24 terms of the list of numbers is 672.
Example 16: A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find: the production in the 1st year, the production in the 10th year, and the total production in the first 7 years.
Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in the 1st, 2nd, 3rd, ... years will form an AP.
Let us denote the number of TV sets manufactured in the nth year by aₙ.
Then, a₃ equals 600 and a₇ equals 700.
So, a plus 2d equals 600, and a plus 6d equals 700.
Solving these equations, we get d equals 25 and a equals 550.
Therefore, production of TV sets in the first year is 550.
Now, a₁₀ equals a plus 9d, which equals 550 plus 9 times 25, which equals 550 plus 225, which equals 775.
So, production of TV sets in the 10th year is 775.
Also, S₇ equals 7/2 times [2 times 550 plus (7 minus 1) times 25], which equals 7/2 times [1100 plus 150], which equals 7/2 times 1250, which equals 4375.
Thus, the total production of TV sets in the first 7 years is 4375.
Now, students, let me summarize what we have learned in this chapter.
First, we learned that an arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term. The fixed number d is called the common difference. The general form of an AP is a, a plus d, a plus 2d, a plus 3d, and so on.
Second, we learned that a given list of numbers a₁, a₂, a₃, ... is an AP if the differences a₂ minus a₁, a₃ minus a₂, a₄ minus a₃, ..., give the same value, that is, if aₖ₊₁ minus aₖ is the same for different values of k.
Third, we learned that in an AP with first term a and common difference d, the nth term (or the general term) is given by aₙ equals a plus (n minus 1) times d.
Fourth, we learned that the sum of the first n terms of an AP is given by S equals n/2 times [2a plus (n minus 1) times d].
Fifth, we learned that if l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by S equals n/2 times (a plus l).
There is also an important concept called the arithmetic mean. If a, b, c are in AP, then b equals (a plus c) divided by 2, and b is called the arithmetic mean of a and c. This is something you might encounter in higher classes.
Students, this chapter is very important because the concepts of arithmetic progressions are used in many different areas of mathematics and in real life. Whether you are calculating simple interest, planning savings, or analyzing any situation where quantities change by a fixed amount each time, understanding arithmetic progressions will help you solve problems easily and efficiently.
In this lesson, we have covered all the main concepts from the chapter - what an AP is, how to identify an AP, how to find the common difference, the formula for the nth term, and the formula for the sum of n terms. We have also worked through many examples to show you how to apply these formulas. With practice, you will become very comfortable with these concepts.
Thank you for listening attentively. Keep practicing, and you will master this chapter in no time!