Welcome dear students! Today we are going to learn about Arithmetic Progressions from Class 10 Maths.
You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone, and so on. We now look for some patterns which occur in our day-to-day life. Let us consider some examples. First, 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. Her salary in ₹ for the 1st, 2nd, 3rd years will be, respectively, 8000, 8500, 9000, and so on. Second, the lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top. In this figure, we see a ladder with horizontal rungs. The bottom rung is 45 cm in length. The lengths in cm of the 1st, 2nd, 3rd, up to the 8th rung from the bottom to the top are, respectively, 45, 43, 41, 39, 37, 35, 33, 31. Third, in a savings scheme, the amount becomes 5/4 times of itself after every 3 years. The maturity amount in ₹ of an investment of ₹ 8000 after 3, 6, 9 and 12 years will be, respectively, 10000, 12500, 15625, 19531.25. Fourth, the number of unit squares in squares with side 1, 2, 3, units are, respectively, 1², 2², 3², and so on. In this figure, we see a sequence of squares with increasing side lengths, each divided into unit squares. Fifth, Shakila puts ₹ 100 into her daughter’s money box when she was one year old and increased the amount by ₹ 50 every year. The amounts of money in ₹ in the box on the 1st, 2nd, 3rd, 4th birthday were 100, 150, 200, 250, respectively. Sixth, a pair of rabbits are too young to produce in their first month. In the second, 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. In this figure, we see a branching diagram representing rabbit pairs over months. Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd, up to the 6th month, respectively are 1, 1, 2, 3, 5, 8.
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In the examples above, we observe some patterns. In some, we find that the succeeding terms are obtained by adding a fixed number, in other by multiplying with a fixed number, in another we find that 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. Let us move on to the formal definition. Consider the following lists of numbers. First list is 1, 2, 3, 4, and so on. Second list is 100, 70, 40, 10, and so on. Third list is -3, -2, -1, 0, and so on. Fourth list is 3, 3, 3, 3, and so on. Fifth list is -1.0, -1.5, -2.0, -2.5, and so on. Each of the numbers in the list is called a term. Given a term, can you write the next term in each of the lists above? If so, how will you write it? Perhaps by following a pattern or rule. Let us observe and write the rule. In the first list, each term is 1 more than the term preceding it. In the second list, each term is 30 less than the term preceding it. In the third list, each term is obtained by adding 1 to the term preceding it. In the fourth list, all the terms in the list are 3, that is, each term is obtained by adding or subtracting 0 to the term preceding it. In the fifth list, each term is obtained by adding -0.5 to the term preceding it. In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression, abbreviated as AP. So, 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.
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Let us denote the first term of an AP by a₁, second term by a₂, nth term by aₙ and the common difference by d. Then the AP becomes a₁, a₂, a₃, and so on, up to aₙ. So, a₂ - a₁ = a₃ - a₂ = ... = aₙ - aₙ₋₁ = d. Some more examples of AP are given. First, the heights in cm of some students of a school standing in a queue in the morning assembly are 147, 148, 149, up to 157. Second, the minimum temperatures in degree celsius 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. Third, the balance money in ₹ after paying 5% of the total loan of ₹ 1000 every month is 950, 900, 850, 800, down to 50. Fourth, the cash prizes in ₹ given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350, up to 750. Fifth, the total savings in ₹ after every month for 10 months when ₹ 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500. It is left as an exercise for you to explain why each of the lists above is an AP. You can see that a, a + d, a + 2d, a + 3d, and so on represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP. Note that in the first to fifth examples above, there are only a finite number of terms. Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions has a last term. The APs in the first to fifth lists in this section, are not finite APs and so they are called infinite Arithmetic Progressions. Such APs do not have a last term. Now, 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 will 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 so on. If a is 6 and d is -3, then the AP is 6, 3, 0, -3, and so on. Similarly, when a = -7 and d = -2, the AP is -7, -9, -11, -13, and so on. When a = 1.0 and d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3, and so on. When a = 0 and d = 1 1/2, the AP is 0, 1 1/2, 3, 4 1/2, 6, and so on. When a = 2 and d = 0, the AP is 2, 2, 2, 2, and so on. 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 is found by subtracting any term from its succeeding term, that is, the term which immediately follows it should be same for an AP. For example, for the list of numbers 6, 9, 12, 15, and so on, we have a₂ - a₁ = 9 - 6 = 3. a₃ - a₂ = 12 - 9 = 3. a₄ - a₃ = 15 - 12 = 3. Here the difference of any two consecutive terms in each case 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, and so on, a₂ - a₁ = 3 - 6 = -3. a₃ - a₂ = 0 - 3 = -3. a₄ - a₃ = -3 - 0 = -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, and so on, up to aₙ, we have d = aₖ₊₁ - 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 - a₁, a₃ - a₂, a₄ - a₃, and so on. It is enough to find only one of them. Consider the list of numbers 1, 1, 2, 3, 5, and so on. 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, and so on, 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.
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Let us make the concept more clear through some examples. Example 1: For the AP 3/2, 1/2, -1/2, -3/2, and so on, write the first term a and the common difference d. Solution: Here, a = 3/2, d = 1/2 - 3/2 = -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, and so on. Solution: We have a₂ - a₁ = 10 - 4 = 6. a₃ - a₂ = 16 - 10 = 6. a₄ - a₃ = 22 - 16 = 6. That is, aₖ₊₁ - 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 + 6 = 28 and 28 + 6 = 34. Second list: 1, -1, -3, -5, and so on. Solution: a₂ - a₁ = -1 - 1 = -2. a₃ - a₂ = -3 - (-1) = -3 + 1 = -2. a₄ - a₃ = -5 - (-3) = -5 + 3 = -2. That is, aₖ₊₁ - 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 + (-2) = -7 and -7 + (-2) = -9. Third list: -2, 2, -2, 2, -2, and so on. Solution: a₂ - a₁ = 2 - (-2) = 2 + 2 = 4. a₃ - a₂ = -2 - 2 = -4. As a₂ - a₁ ≠ a₃ - a₂, the given list of numbers does not form an AP. Fourth list: 1, 1, 1, 2, 2, 2, 3, 3, 3, and so on. Solution: a₂ - a₁ = 1 - 1 = 0. a₃ - a₂ = 1 - 1 = 0. a₄ - a₃ = 2 - 1 = 1. Here, a₂ - a₁ = a₃ - a₂ but is not equal to a₄ - a₃. So, the given list of numbers does not form an AP.
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Now let us solve Exercise 5.1 completely. Question 1: In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? First situation: The taxi fare after each kilometre when the fare is ₹ 15 for the first kilometre and ₹ 8 for each additional kilometre. Solution: The fares are 15, 15 + 8 = 23, 23 + 8 = 31, and so on. The difference between consecutive terms is 23 - 15 = 8, and 31 - 23 = 8. Since the difference is constant, it forms an AP. Second situation: The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. Solution: Let initial air be 1. After first removal, remaining is 1 - 1/4 = 3/4. After second removal, remaining is 3/4 - 1/4 of 3/4 = 3/4 - 3/16 = 9/16. Differences are 3/4 - 1 = -1/4, and 9/16 - 3/4 = -3/16. Since differences are not equal, it does not form an AP. Third situation: The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre. Solution: Costs are 150, 200, 250, 300, and so on. Difference is 200 - 150 = 50, 250 - 200 = 50. Constant difference, so it forms an AP. Fourth situation: The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum. Solution: Amounts are 10000 × 1.08, 10000 × 1.08², 10000 × 1.08³. Differences are not constant. So it does not form an AP. Question 2: Write first four terms of the AP, when the first term a and the common difference d are given as follows. First: a = 10, d = 10. Terms are 10, 20, 30, 40. Second: a = -2, d = 0. Terms are -2, -2, -2, -2. Third: a = 4, d = -3. Terms are 4, 1, -2, -5. Fourth: a = -1, d = 1/2. Terms are -1, -1/2, 0, 1/2. Fifth: a = -1.25, d = -0.25. Terms are -1.25, -1.50, -1.75, -2.00. Question 3: For the following APs, write the first term and the common difference. First: 3, 1, -1, -3, and so on. a = 3, d = 1 - 3 = -2. Second: -5, -1, 3, 7, and so on. a = -5, d = -1 - (-5) = 4. Third: 1/3, 5/3, 9/3, 13/3, and so on. a = 1/3, d = 5/3 - 1/3 = 4/3. Fourth: 0.6, 1.7, 2.8, 3.9, and so on. a = 0.6, d = 1.7 - 0.6 = 1.1. Question 4: Which of the following are APs? If they form an AP, find the common difference d and write three more terms. First: 2, 4, 8, 16, and so on. Differences are 2, 4, 8. Not constant. Not an AP. Second: 2, 5/2, 3, 7/2, and so on. Differences are 1/2, 1/2, 1/2. d = 1/2. Next terms: 4, 9/2, 5. Third: -1.2, -3.2, -5.2, -7.2, and so on. Differences are -2, -2, -2. d = -2. Next terms: -9.2, -11.2, -13.2. Fourth: -10, -6, -2, 2, and so on. Differences are 4, 4, 4. d = 4. Next terms: 6, 10, 14. Fifth: 3, 3 + √2, 3 + 2√2, 3 + 3√2, and so on. Differences are √2, √2, √2. d = √2. Next terms: 3 + 4√2, 3 + 5√2, 3 + 6√2. Sixth: 0.2, 0.22, 0.222, 0.2222, and so on. Differences are 0.02, 0.002, 0.0002. Not constant. Not an AP. Seventh: 0, -4, -8, -12, and so on. Differences are -4, -4, -4. d = -4. Next terms: -16, -20, -24. Eighth: -1/2, -1/2, -1/2, -1/2, and so on. Differences are 0, 0, 0. d = 0. Next terms: -1/2, -1/2, -1/2. Ninth: 1, 3, 9, 27, and so on. Differences are 2, 6, 18. Not constant. Not an AP. Tenth: a, 2a, 3a, 4a, and so on. Differences are a, a, a. d = a. Next terms: 5a, 6a, 7a. Eleventh: a, a², a³, a⁴, and so on. Differences are a² - a, a³ - a². Not constant unless a is 0 or 1. Generally not an AP. Twelfth: √2, √8, √18, √32, and so on. These simplify to √2, 2√2, 3√2, 4√2. Differences are √2, √2, √2. d = √2. Next terms: 5√2, 6√2, 7√2. Thirteenth: √3, √6, √9, √12, and so on. Differences are not constant. Not an AP. Fourteenth: 1², 3², 5², 7², and so on. That is 1, 9, 25, 49. Differences are 8, 16, 24. Not constant. Not an AP. Fifteenth: 1², 5², 7², 7³, and so on. That is 1, 25, 49, 343. Differences are not constant. Not an AP.
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Let us move to Section 5.3, nth Term of an AP. Let us consider the situation again, given in Section 5.1 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 + 500 = ₹ 8500. In the same way, we can find the monthly salary for the third, fourth and fifth year by adding ₹ 500 to the salary of the previous year. So, the salary for the third year = 8500 + 500 = 8000 + 500 + 500 = 8000 + 2 × 500 = 8000 + (3 - 1) × 500, which is ₹ 9000. Salary for the fourth year = 9000 + 500 = 8000 + 500 + 500 + 500 = 8000 + 3 × 500 = 8000 + (4 - 1) × 500, which is ₹ 9500. Salary for the fifth year = 9500 + 500 = 8000 + 500 + 500 + 500 + 500 = 8000 + 4 × 500 = 8000 + (5 - 1) × 500, which is ₹ 10000. Observe that we are getting a list of numbers 8000, 8500, 9000, 9500, 10000, and so on. These numbers are in AP. Now, looking at the pattern formed above, can you find her monthly salary for the sixth year? The fifteenth year? And, assuming that she will still be working in the job, what about the monthly salary for the twenty-fifth year? You would calculate this by adding ₹ 500 each time to the salary of the previous year to give the answer. Can we make this process shorter? Let us see. Salary for the fifteenth year = Salary for the fourteenth year + ₹ 500 = 8000 + 14 × 500 = 8000 + (15 - 1) × 500 = ₹ 15000. That is, First salary + (15 - 1) × Annual increment. In the same way, her monthly salary for the twenty-fifth year would be 8000 + (25 - 1) × 500 = ₹ 20000. This example would have given you some idea about how to write the fifteenth term, or the twenty-fifth term, and more generally, the nth term of the AP. Let a₁, a₂, a₃, and so on be an AP whose first term a₁ is a and the common difference is d. Then, the second term a₂ = a + d = a + (2 - 1)d. The third term a₃ = a₂ + d = (a + d) + d = a + 2d = a + (3 - 1)d. The fourth term a₄ = a₃ + d = (a + 2d) + d = a + 3d = a + (4 - 1)d. Looking at the pattern, we can say that the nth term aₙ = a + (n - 1)d. So, the nth term aₙ of the AP with first term a and common difference d is given by aₙ = a + (n - 1)d. 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.
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Let us consider some examples. Example 3: Find the 10th term of the AP 2, 7, 12, and so on. Solution: Here, a = 2, d = 7 - 2 = 5 and n = 10. We have aₙ = a + (n - 1)d. So, a₁₀ = 2 + (10 - 1) × 5 = 2 + 45 = 47. Therefore, the 10th term of the given AP is 47. Example 4: Which term of the AP 21, 18, 15, and so on is -81? Also, is any term 0? Give reason for your answer. Solution: Here, a = 21, d = 18 - 21 = -3 and aₙ = -81, and we have to find n. As aₙ = a + (n - 1)d, we have -81 = 21 + (n - 1)(-3). This gives -81 = 24 - 3n. So, -105 = -3n. Thus, n = 35. Therefore, the 35th term of the given AP is -81. Next, we want to know if there is any n for which aₙ = 0. If such an n is there, then 21 + (n - 1)(-3) = 0, that is, 3(n - 1) = 21. That is, n = 8. So, the eighth term is 0. Example 5: Determine the AP whose 3rd term is 5 and the 7th term is 9. Solution: We have a₃ = a + (3 - 1)d = a + 2d = 5. And a₇ = a + (7 - 1)d = a + 6d = 9. Solving the pair of linear equations, we subtract the first from the second to get 4d = 4, so d = 1. Substituting d = 1 into a + 2d = 5 gives a + 2 = 5, so a = 3. Hence, the required AP is 3, 4, 5, 6, 7, and so on. Example 6: Check whether 301 is a term of the list of numbers 5, 11, 17, 23, and so on. Solution: We have a₂ - a₁ = 11 - 5 = 6. a₃ - a₂ = 17 - 11 = 6. a₄ - a₃ = 23 - 17 = 6. As aₖ₊₁ - aₖ is the same for k = 1, 2, 3, and so on, the given list of numbers is an AP. Now, a = 5 and d = 6. Let 301 be a term, say, the nth term of this AP. We know that aₙ = a + (n - 1)d. So, 301 = 5 + (n - 1) × 6. That is, 301 = 6n - 1. So, n = 302/6 = 151/3. But n should be a positive integer. So, 301 is not a term of the given list of numbers. Example 7: How many two-digit numbers are divisible by 3? Solution: The list of two-digit numbers divisible by 3 is 12, 15, 18, and so on, up to 99. Is this an AP? Yes it is. Here, a = 12, d = 3, aₙ = 99. As aₙ = a + (n - 1)d, we have 99 = 12 + (n - 1) × 3. That is, 87 = (n - 1) × 3. That is, n - 1 = 87/3 = 29. That is, n = 29 + 1 = 30. So, 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, and so on, up to -62. Solution: Here, a = 10, d = 7 - 10 = -3, l = -62, where l = a + (n - 1)d. To find the 11th term from the last term, we will find the total number of terms in the AP. So, -62 = 10 + (n - 1)(-3). That is, -72 = (n - 1)(-3). That is, n - 1 = 24 or n = 25. So, there are 25 terms in the given AP. The 11th term from the last term will be the 15th term. So, a₁₅ = 10 + (15 - 1)(-3) = 10 - 42 = -32. That is, the 11th term from the last term is -32. Alternative Solution: If we write the given AP in the reverse order, then a = -62 and d = 3. So, the question now becomes finding the 11th term with these a and d. So, a₁₁ = -62 + (11 - 1) × 3 = -62 + 30 = -32. So, the 11th term, which is now the required term, is -32.
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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. Solution: We know that the formula to calculate simple interest is given by Simple Interest = (P × R × T) / 100. So, the interest at the end of the first year = (1000 × 8 × 1) / 100 ₹ = ₹ 80. The interest at the end of the second year = (1000 × 8 × 2) / 100 ₹ = ₹ 160. The interest at the end of the third year = (1000 × 8 × 3) / 100 ₹ = ₹ 240. Similarly, we can obtain the interest at the end of the fourth year, fifth year, and so on. So, the interest in ₹ at the end of the first, second, third, and so on years, respectively are 80, 160, 240, and so on. It is an AP as the difference between the consecutive terms in the list is 80, that is, d = 80. Also, a = 80. So, to find the interest at the end of 30 years, we shall find a₃₀. Now, a₃₀ = a + (30 - 1)d = 80 + 29 × 80 = 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? Solution: The number of rose plants in the first, second, third, and so on rows are 23, 21, 19, and so on, up to 5. It forms an AP because the difference between consecutive terms is -2. Let the number of rows in the flower bed be n. Then a = 23, d = 21 - 23 = -2, aₙ = 5. As aₙ = a + (n - 1)d, we have 5 = 23 + (n - 1)(-2). That is, -18 = (n - 1)(-2). That is, n = 10. So, there are 10 rows in the flower bed.
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Now let us solve Exercise 5.2 completely. Question 1: 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. First row: a = 7, d = 3, n = 8. aₙ = a + (n - 1)d = 7 + 7 × 3 = 28. Second row: a = -18, n = 10, aₙ = 0. 0 = -18 + 9d. 18 = 9d, so d = 2. Third row: d = -3, n = 18, aₙ = -5. -5 = a + 17 × (-3). -5 = a - 51, so a = 46. Fourth row: a = -18.9, d = 2.5, aₙ = 3.6. 3.6 = -18.9 + (n - 1) × 2.5. 22.5 = (n - 1) × 2.5. n - 1 = 9, so n = 10. Fifth row: a = 3.5, d = 0, n = 105. aₙ = 3.5 + 104 × 0 = 3.5. Question 2: Choose the correct choice and justify. First: 30th term of the AP 10, 7, 4, and so on. a = 10, d = -3. a₃₀ = 10 + 29 × (-3) = 10 - 87 = -77. Correct choice is C. Second: 11th term of the AP -3, -1/2, 2, and so on. a = -3, d = -1/2 - (-3) = 5/2. a₁₁ = -3 + 10 × 5/2 = -3 + 25 = 22. Correct choice is B. Question 3: Find missing terms. First: 2, blank, 26. Let middle term be x. x - 2 = 26 - x. 2x = 28, x = 14. Second: blank, 13, blank, 3. Let terms be a, 13, b, 3. d = 13 - a. b = 13 + d. 3 = b + d. Solving gives d = -5. So a = 18, b = 8. Sequence: 18, 13, 8, 3. Third: 5, blank, blank, 1/2. Let terms be 5, a, b, 1/2. d = a - 5. b = a + d. 1/2 = b + d. Solving gives d = -3/2. a = 7/2, b = 2. Sequence: 5, 7/2, 2, 1/2. Fourth: -4, blank, blank, blank, blank, 6. Six terms total. a₁ = -4, a₆ = 6. 6 = -4 + 5d. 10 = 5d, d = 2. Missing terms: -2, 0, 2, 4. Fifth: blank, 38, blank, blank, blank, -22. Six terms. a₂ = 38, a₆ = -22. -22 = 38 + 4d. 4d = -60, d = -15. a₁ = 53. Missing terms: 23, 8, -7. Sequence: 53, 38, 23, 8, -7, -22. Question 4: Which term of the AP 3, 8, 13, 18, and so on is 78? a = 3, d = 5. 78 = 3 + (n - 1) × 5. 75 = 5 × (n - 1). n - 1 = 15, n = 16. The 16th term is 78. Question 5: Find number of terms. First: 7, 13, 19, and so on, up to 205. a = 7, d = 6. 205 = 7 + (n - 1) × 6. 198 = 6 × (n - 1). n - 1 = 33, n = 34. Second: 18, 15 1/2, 13, and so on, up to -47. a = 18, d = -5/2. -47 = 18 + (n - 1) × (-5/2). -65 = (n - 1) × (-5/2). n - 1 = 26, n = 27. Question 6: Check whether -150 is a term of the AP 11, 8, 5, 2, and so on. a = 11, d = -3. -150 = 11 + (n - 1) × (-3). -161 = -3 × (n - 1). n - 1 = 161/3, which is not an integer. So -150 is not a term. Question 7: Find 31st term of AP whose 11th term is 38 and 16th term is 73. a₁₁ = a + 10d = 38. a₁₆ = a + 15d = 73. Subtracting gives 5d = 35, so d = 7. a + 70 = 38, so a = -32. a₃₁ = -32 + 30 × 7 = -32 + 210 = 178. Question 8: AP has 50 terms. 3rd term is 12, last term is 106. Find 29th term. a₃ = a + 2d = 12. a₅₀ = a + 49d = 106. Subtracting gives 47d = 94, so d = 2. a + 4 = 12, so a = 8. a₂₉ = 8 + 28 × 2 = 8 + 56 = 64. Question 9: 3rd term is 4, 9th term is -8. Which term is zero? a + 2d = 4. a + 8d = -8. Subtracting gives 6d = -12, so d = -2. a - 4 = 4, so a = 8. Set aₙ = 0. 8 + (n - 1)(-2) = 0. 8 = 2 × (n - 1). n - 1 = 4, n = 5. The 5th term is zero. Question 10: 17th term exceeds 10th term by 7. Find d. a₁₇ - a₁₀ = 7. (a + 16d) - (a + 9d) = 7. 7d = 7, so d = 1. Question 11: Which term of AP 3, 15, 27, 39, and so on will be 132 more than its 54th term? a = 3, d = 12. a₅₄ = 3 + 53 × 12 = 3 + 636 = 639. We need aₙ = 639 + 132 = 771. 771 = 3 + (n - 1) × 12. 768 = 12 × (n - 1). n - 1 = 64, n = 65. The 65th term. Question 12: Two APs have same common difference. Difference between 100th terms is 100. Difference between 1000th terms? Let first terms be a₁ and a₂, common difference d. a₁₀₀ first - a₁₀₀ second = (a₁ + 99d) - (a₂ + 99d) = a₁ - a₂ = 100. a₁₀₀₀ first - a₁₀₀₀ second = (a₁ + 999d) - (a₂ + 999d) = a₁ - a₂ = 100. Question 13: How many three-digit numbers divisible by 7? First is 105, last is 994. a = 105, d = 7, aₙ = 994. 994 = 105 + (n - 1) × 7. 889 = 7 × (n - 1). n - 1 = 127, n = 128. Question 14: How many multiples of 4 lie between 10 and 250? First is 12, last is 248. a = 12, d = 4, aₙ = 248. 248 = 12 + (n - 1) × 4. 236 = 4 × (n - 1). n - 1 = 59, n = 60. Question 15: For what value of n are nth terms of 63, 65, 67, and so on and 3, 10, 17, and so on equal? First AP: a = 63, d = 2. aₙ = 63 + 2(n - 1) = 61 + 2n. Second AP: a = 3, d = 7. aₙ = 3 + 7(n - 1) = 7n - 4. Equate: 61 + 2n = 7n - 4. 65 = 5n, n = 13. Question 16: Determine AP whose third term is 16 and 7th term exceeds 5th term by 12. a + 2d = 16. (a + 6d) - (a + 4d) = 12. 2d = 12, d = 6. a + 12 = 16, a = 4. AP is 4, 10, 16, 22, and so on. Question 17: Find 20th term from last term of AP 3, 8, 13, and so on, up to 253. Reverse AP: a = 253, d = -5. a₂₀ = 253 + 19 × (-5) = 253 - 95 = 158. Question 18: Sum of 4th and 8th terms is 24. Sum of 6th and 10th terms is 44. Find first three terms. (a + 3d) + (a + 7d) = 24. 2a + 10d = 24. a + 5d = 12. (a + 5d) + (a + 9d) = 44. 2a + 14d = 44. a + 7d = 22. Subtract: 2d = 10, d = 5. a + 25 = 12, a = -13. First three terms: -13, -8, -3. Question 19: Subba Rao started in 1995 at ₹ 5000, increment ₹ 200 each year. When did income reach ₹ 7000? a = 5000, d = 200. aₙ = 7000. 7000 = 5000 + (n - 1) × 200. 2000 = 200 × (n - 1). n - 1 = 10, n = 11. Year is 1995 + 10 = 2005. Question 20: Ramkali saved ₹ 5 first week, increased by ₹ 1.75. nth week savings ₹ 20.75. Find n. a = 5, d = 1.75. aₙ = 20.75. 20.75 = 5 + (n - 1) × 1.75. 15.75 = 1.75 × (n - 1). n - 1 = 9, n = 10.
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Let us move to Section 5.4, Sum of First n Terms of an AP. Let us consider the situation again given in Section 5.1 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 ₹ put in the money box on her first, second, third, fourth birthday were respectively 100, 150, 200, 250, and so on 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 did he do? He wrote S = 1 + 2 + 3 + ... + 99 + 100. And then, reversed the numbers to write S = 100 + 99 + ... + 3 + 2 + 1. Adding these two, he got 2S = (100 + 1) + (99 + 2) + ... + (3 + 98) + (2 + 99) + (1 + 100) = 101 + 101 + ... + 101 + 101, 100 times. So, S = (100 × 101) / 2 = 5050. We will now use the same technique to find the sum of the first n terms of an AP: a, a + d, a + 2d, and so on. The nth term of this AP is a + (n - 1)d. Let S denote the sum of the first n terms of the AP. We have S = a + (a + d) + (a + 2d) + ... + [a + (n - 1)d]. Rewriting the terms in reverse order, we have S = [a + (n - 1)d] + [a + (n - 2)d] + ... + (a + d) + a. On adding these two equations term-wise, we get 2S = [2a + (n - 1)d] + [2a + (n - 1)d] + ... + [2a + (n - 1)d], n times. So, 2S = n[2a + (n - 1)d]. Thus, S = n/2 [2a + (n - 1)d]. So, the sum of the first n terms of an AP is given by S = n/2 [2a + (n - 1)d]. We can also write this as S = n/2 [a + a + (n - 1)d], that is, S = n/2 (a + aₙ). Now, if there are only n terms in an AP, then aₙ = l, the last term. From this, we see that S = n/2 (a + 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 ₹ in the money box of Shakila’s daughter on first, second, third, fourth birthday, and so on, were 100, 150, 200, 250, and so on, 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 = 100, d = 50 and n = 21. Using the formula S = n/2 [2a + (n - 1)d], we have S = 21/2 [200 + 20 × 50] = 21/2 [200 + 1000] = 21/2 × 1200 = 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 - 1 terms of it, that is, aₙ = Sₙ - Sₙ₋₁.
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Let us consider some examples. Example 11: Find the sum of the first 22 terms of the AP 8, 3, -2, and so on. Solution: Here, a = 8, d = 3 - 8 = -5, n = 22. We know that S = n/2 [2a + (n - 1)d]. Therefore, S = 22/2 [16 + 21 × (-5)] = 11(16 - 105) = 11(-89) = -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. Solution: Here, S₁₄ = 1050, n = 14, a = 10. As Sₙ = n/2 [2a + (n - 1)d], so, 1050 = 14/2 [20 + 13d] = 7(20 + 13d) = 140 + 91d. That is, 910 = 91d or d = 10. Therefore, a₂₀ = 10 + 19 × 10 = 200. That is, the 20th term is 200. Example 13: How many terms of the AP 24, 21, 18, and so on must be taken so that their sum is 78? Solution: Here, a = 24, d = 21 - 24 = -3, Sₙ = 78. We need to find n. We know that Sₙ = n/2 [2a + (n - 1)d]. So, 78 = n/2 [48 + (n - 1)(-3)] = n/2 [51 - 3n]. Multiplying by 2 gives 156 = 51n - 3n². Or 3n² - 51n + 156 = 0. Dividing by 3 gives n² - 17n + 52 = 0. Factoring gives (n - 4)(n - 13) = 0. So n = 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 = the sum of the first 13 terms = 78. Two answers are possible because the sum of the terms from 5th to 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 first 1000 positive integers and the sum of first n positive integers. Solution: First, let S = 1 + 2 + 3 + ... + 1000. Using the formula Sₙ = n/2 (a + l), we have S₁₀₀₀ = 1000/2 (1 + 1000) = 500 × 1001 = 500500. So, the sum of the first 1000 positive integers is 500500. Second, let Sₙ = 1 + 2 + 3 + ... + n. Here a = 1 and the last term l is n. Therefore, Sₙ = n/2 (1 + n). So, the sum of first n positive integers is given by Sₙ = n(n + 1)/2. Example 15: Find the sum of first 24 terms of the list of numbers whose nth term is given by aₙ = 3 + 2n. Solution: As aₙ = 3 + 2n, so a₁ = 3 + 2 = 5. a₂ = 3 + 4 = 7. a₃ = 3 + 6 = 9. List of numbers becomes 5, 7, 9, 11, and so on. Here, 7 - 5 = 9 - 7 = 11 - 9 = 2 and so on. So, it forms an AP with common difference d = 2. To find S₂₄, we have n = 24, a = 5, d = 2. Therefore, S₂₄ = 24/2 [10 + 23 × 2] = 12(10 + 46) = 12 × 56 = 672. So, sum of 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 first year, the production in the tenth year, and the total production in first seven years. Solution: Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in first, second, third, and so on years will form an AP. Let us denote the number of TV sets manufactured in the nth year by aₙ. Then, a₃ = 600 and a₇ = 700. So, a + 2d = 600 and a + 6d = 700. Solving these equations, we get 4d = 100, so d = 25. Substituting d = 25 gives a + 50 = 600, so a = 550. Therefore, production of TV sets in the first year is 550. Second, now a₁₀ = a + 9d = 550 + 9 × 25 = 550 + 225 = 775. So, production of TV sets in the 10th year is 775. Third, also S₇ = 7/2 [1100 + 6 × 25] = 7/2 [1100 + 150] = 7/2 × 1250 = 4375. Thus, the total production of TV sets in first 7 years is 4375.
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Now let us solve Exercise 5.3 completely. Question 1: Find the sum of the following APs. First: 2, 7, 12, and so on, to 10 terms. a = 2, d = 5, n = 10. S₁₀ = 10/2 [4 + 9 × 5] = 5(4 + 45) = 5 × 49 = 245. Second: -37, -33, -29, and so on, to 12 terms. a = -37, d = 4, n = 12. S₁₂ = 12/2 [-74 + 11 × 4] = 6(-74 + 44) = 6 × (-30) = -180. Third: 0.6, 1.7, 2.8, and so on, to 100 terms. a = 0.6, d = 1.1, n = 100. S₁₀₀ = 100/2 [1.2 + 99 × 1.1] = 50(1.2 + 108.9) = 50 × 110.1 = 5505. Fourth: 1/15, 1/12, 1/10, and so on, to 11 terms. a = 1/15, d = 1/12 - 1/15 = 1/60. n = 11. S₁₁ = 11/2 [2/15 + 10 × 1/60] = 11/2 [2/15 + 1/6] = 11/2 [4/30 + 5/30] = 11/2 × 9/30 = 11/2 × 3/10 = 33/20. Question 2: Find the sums given below. First: 7 + 10 1/2 + 14 + ... + 84. a = 7, d = 3 1/2. Last term l = 84. 84 = 7 + (n - 1) × 3 1/2. 77 = (n - 1) × 3 1/2. n - 1 = 22, n = 23. S₂₃ = 23/2 (7 + 84) = 23/2 × 91 = 2093/2 = 1046.5. Second: 34 + 32 + 30 + ... + 10. a = 34, d = -2, l = 10. 10 = 34 + (n - 1) × (-2). -24 = -2 × (n - 1). n - 1 = 12, n = 13. S₁₃ = 13/2 (34 + 10) = 13/2 × 44 = 13 × 22 = 286. Third: -5 + (-8) + (-11) + ... + (-230). a = -5, d = -3, l = -230. -230 = -5 + (n - 1) × (-3). -225 = -3 × (n - 1). n - 1 = 75, n = 76. S₇₆ = 76/2 (-5 + (-230)) = 38 × (-235) = -8930. Question 3: In an AP. First: a = 5, d = 3, aₙ = 50. Find n and Sₙ. 50 = 5 + (n - 1) × 3. 45 = 3 × (n - 1). n - 1 = 15, n = 16. S₁₆ = 16/2 (5 + 50) = 8 × 55 = 440. Second: a = 7, a₁₃ = 35. Find d and S₁₃. 35 = 7 + 12d. 28 = 12d, d = 7/3. S₁₃ = 13/2 (7 + 35) = 13/2 × 42 = 13 × 21 = 273. Third: a₁₂ = 37, d = 3. Find a and S₁₂. 37 = a + 11 × 3. a = 37 - 33 = 4. S₁₂ = 12/2 (4 + 37) = 6 × 41 = 246. Fourth: a₃ = 15, S₁₀ = 125. Find d and a₁₀. a + 2d = 15. 10/2 [2a + 9d] = 125. 2a + 9d = 25. Multiply first by 2: 2a + 4d = 30. Subtract: 5d = -5, d = -1. a - 2 = 15, a = 17. a₁₀ = 17 + 9 × (-1) = 8. Fifth: d = 5, S₉ = 75. Find a and a₉. 9/2 [2a + 8 × 5] = 75. 9/2 [2a + 40] = 75. 2a + 40 = 150/9 = 50/3. 2a = 50/3 - 120/3 = -70/3. a = -35/3. a₉ = -35/3 + 8 × 5 = -35/3 + 120/3 = 85/3. Sixth: a = 2, d = 8, Sₙ = 90. Find n and aₙ. n/2 [4 + 8(n - 1)] = 90. n(8n - 4) = 180. 8n² - 4n - 180 = 0. 2n² - n - 45 = 0. 2n² - 10n + 9n - 45 = 0. 2n(n - 5) + 9(n - 5) = 0. n = 5. a₅ = 2 + 4 × 8 = 34. Seventh: a = 8, aₙ = 62, Sₙ = 210. Find n and d. 210 = n/2 (8 + 62) = n/2 × 70 = 35n. n = 6. 62 = 8 + 5d. 5d = 54, d = 54/5. Eighth: aₙ = 4, d = 2, Sₙ = -14. Find n and a. -14 = n/2 [2a + 2(n - 1)]. -28 = n[2a + 2n - 2]. Also 4 = a + 2(n - 1), so a = 6 - 2n. Substitute: -28 = n[12 - 4n + 2n - 2] = n[10 - 2n]. 2n² - 10n - 28 = 0. n² - 5n - 14 = 0. (n - 7)(n + 2) = 0. n = 7. a = 6 - 14 = -8. Ninth: a = 3, n = 8, S = 192. Find d. 192 = 8/2 [6 + 7d]. 192 = 4(6 + 7d). 48 = 6 + 7d. 7d = 42, d = 6. Tenth: l = 28, S = 144, n = 9. Find a. 144 = 9/2 (a + 28). 32 = a + 28. a = 4.
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Question 4: How many terms of AP 9, 17, 25, and so on must be taken to give sum 636? a = 9, d = 8. n/2 [18 + 8(n - 1)] = 636. n(10 + 8n) = 1272. 8n² + 10n - 1272 = 0. 4n² + 5n - 636 = 0. Solving gives n = 12. Question 5: First term 5, last term 45, sum 400. Find n and d. 400 = n/2 (5 + 45) = 25n. n = 16. 45 = 5 + 15d. 15d = 40, d = 8/3. Question 6: First 17, last 350, d = 9. Find n and sum. 350 = 17 + (n - 1) × 9. 333 = 9 × (n - 1). n - 1 = 37, n = 38. S₃₈ = 38/2 (17 + 350) = 19 × 367 = 6973. Question 7: Sum of first 22 terms, d = 7, 22nd term 149. Find sum. a₂₂ = a + 21 × 7 = 149. a + 147 = 149, a = 2. S₂₂ = 22/2 (2 + 149) = 11 × 151 = 1661. Question 8: Sum of first 51 terms, second term 14, third term 18. Find sum. a + d = 14. a + 2d = 18. d = 4, a = 10. S₅₁ = 51/2 [20 + 50 × 4] = 51/2 × 220 = 51 × 110 = 5610. Question 9: Sum of first 7 terms 49, sum of 17 terms 289. Find sum of first n terms. S₇ = 7/2 [2a + 6d] = 49. 2a + 6d = 14. a + 3d = 7. S₁₇ = 17/2 [2a + 16d] = 289. 2a + 16d = 34. a + 8d = 17. Subtract: 5d = 10, d = 2. a + 6 = 7, a = 1. Sₙ = n/2 [2 + 2(n - 1)] = n². Question 10: Show that sequences form AP and find sum of first 15 terms. First: aₙ = 3 + 4n. a₁ = 7, a₂ = 11, a₃ = 15. d = 4. It is an AP. S₁₅ = 15/2 [14 + 14 × 4] = 15/2 × 70 = 525. Second: aₙ = 9 - 5n. a₁ = 4, a₂ = -1, a₃ = -6. d = -5. It is an AP. S₁₅ = 15/2 [8 + 14 × (-5)] = 15/2 × (-62) = -465. Question 11: Sum of first n terms is 4n - n². Find first term, sum of first two, second term, third, tenth, nth. S₁ = 4 - 1 = 3. First term is 3. S₂ = 8 - 4 = 4. Second term = S₂ - S₁ = 4 - 3 = 1. S₃ = 12 - 9 = 3. Third term = 3 - 4 = -1. S₁₀ = 40 - 100 = -60. S₉ = 36 - 81 = -45. Tenth term = -15. aₙ = Sₙ - Sₙ₋₁ = 4n - n² - [4(n - 1) - (n - 1)²] = 5 - 2n. Question 12: Sum of first 40 positive integers divisible by 6. AP: 6, 12, 18, and so on. a = 6, d = 6, n = 40. S₄₀ = 40/2 [12 + 39 × 6] = 20(12 + 234) = 20 × 246 = 4920. Question 13: Sum of first 15 multiples of 8. AP: 8, 16, 24, and so on. a = 8, d = 8, n = 15. S₁₅ = 15/2 [16 + 14 × 8] = 15/2 × 128 = 15 × 64 = 960. Question 14: Sum of odd numbers between 0 and 50. AP: 1, 3, 5, and so on, up to 49. a = 1, d = 2, l = 49. 49 = 1 + (n - 1) × 2. n = 25. S₂₅ = 25/2 (1 + 49) = 25 × 25 = 625. Question 15: Penalty ₹ 200 first day, ₹ 250 second, ₹ 300 third, and so on. Delay 30 days. Find total. a = 200, d = 50, n = 30. S₃₀ = 30/2 [400 + 29 × 50] = 15(400 + 1450) = 15 × 1850 = ₹ 27750. Question 16: Sum ₹ 700 for seven prizes. Each prize ₹ 20 less than preceding. Find values. a, a - 20, a - 40, and so on. S₇ = 7/2 [2a + 6 × (-20)] = 700. 2a - 120 = 200. 2a = 320, a = 160. Prizes: 160, 140, 120, 100, 80, 60, 40. Question 17: Trees planted. Class 1 plants 1, Class 2 plants 2, up to Class 12. Three sections each. Total trees? AP: 3, 6, 9, and so on, up to 36. a = 3, d = 3, n = 12. S₁₂ = 12/2 [6 + 11 × 3] = 6 × 39 = 234. Question 18: Spiral of semicircles, radii 0.5, 1.0, 1.5, 2.0, and so on, up to 13 semicircles. Total length? Length of semicircle is πr. Radii form AP with a = 0.5, d = 0.5. Lengths form AP with a = 0.5π, d = 0.5π, n = 13. S₁₃ = 13/2 [π + 12 × 0.5π] = 13/2 × 7π = 91π/2. Using π = 22/7, length = 91 × 11 / 2 = 1001/2 = 500.5 cm. Question 19: 200 logs stacked. 20 bottom, 19 next, 18 next, and so on. Find rows and top logs. AP: 20, 19, 18, and so on. a = 20, d = -1. Sₙ = 200. n/2 [40 - (n - 1)] = 200. 41n - n² = 400. n² - 41n + 400 = 0. (n - 16)(n - 25) = 0. n = 16 or 25. If n = 16, a₁₆ = 20 - 15 = 5. If n = 25, a₂₅ = 20 - 24 = -4. Not possible. So 16 rows, 5 logs on top. Question 20: Potato race. Bucket 5 m from first potato, others 3 m apart. 10 potatoes. Total distance? Distances run: 2 × 5, 2 × 8, 2 × 11, and so on, 10 terms. AP: 10, 16, 22, and so on. a = 10, d = 6, n = 10. S₁₀ = 10/2 [20 + 9 × 6] = 5(20 + 54) = 5 × 74 = 370 metres.
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Now let us solve Exercise 5.4, which is optional. Question 1: Which term of AP 121, 117, 113, and so on is its first negative term? a = 121, d = -4. aₙ = 121 - 4(n - 1) = 125 - 4n. Set aₙ < 0. 125 - 4n < 0. 4n > 125. n > 31.25. First integer is 32. The 32nd term is the first negative term. Question 2: Sum of third and seventh terms is 6, product is 8. Find sum of first sixteen terms. (a + 2d) + (a + 6d) = 6. 2a + 8d = 6. a + 4d = 3. a = 3 - 4d. Product: (a + 2d)(a + 6d) = 8. Substitute a: (3 - 2d)(3 + 2d) = 8. 9 - 4d² = 8. 4d² = 1. d² = 1/4. d = 1/2 or -1/2. If d = 1/2, a = 1. S₁₆ = 16/2 [2 + 15 × 1/2] = 8(2 + 7.5) = 8 × 9.5 = 76. If d = -1/2, a = 5. S₁₆ = 16/2 [10 + 15 × (-1/2)] = 8(10 - 7.5) = 8 × 2.5 = 20. Question 3: Ladder rungs 25 cm apart. Lengths decrease from 45 cm at bottom to 25 cm at top. Top and bottom rungs 2.5 m apart. Find wood length. Distance between rungs is 25 cm. Total distance 250 cm. Number of rungs = 250/25 + 1 = 11. AP of lengths: 45, 43, 41, and so on, up to 25. a = 45, d = -2, n = 11. S₁₁ = 11/2 (45 + 25) = 11/2 × 70 = 385 cm. Question 4: Houses numbered 1 to 49. Find x such that sum before x equals sum after x. Sum 1 to x - 1 = Sum x + 1 to 49. (x - 1)x/2 = 49 × 50/2 - x(x + 1)/2. Multiply by 2: x² - x = 2450 - x² - x. 2x² = 2450. x² = 1225. x = 35. Question 5: Terrace 15 steps, each 50 m long. Rise 1/4 m, tread 1/2 m. Find concrete volume. Volume of first step = 50 × 1/4 × 1/2. Second step: 50 × 2/4 × 1/2. Third: 50 × 3/4 × 1/2. AP of volumes: 50/8, 100/8, 150/8, and so on. a = 50/8, d = 50/8, n = 15. S₁₅ = 15/2 [100/8 + 14 × 50/8] = 15/2 [(100 + 700)/8] = 15/2 × 800/8 = 15/2 × 100 = 750 cubic metres.
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Let us conclude with the Summary. In this chapter, you have studied the following points. First, an arithmetic progression 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 + d, a + 2d, a + 3d, and so on. Second, a given list of numbers a₁, a₂, a₃, and so on is an AP, if the differences a₂ - a₁, a₃ - a₂, a₄ - a₃, and so on, give the same value, that is, if aₖ₊₁ - aₖ is the same for different values of k. Third, in an AP with first term a and common difference d, the nth term or the general term is given by aₙ = a + (n - 1)d. Fourth, the sum of the first n terms of an AP is given by S = n/2 [2a + (n - 1)d]. Fifth, 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 = n/2 (a + l). A note to the reader: If a, b, c are in AP, then b = (a + c)/2 and b is called the arithmetic mean of a and c.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]