Worksheet
A chapter that explores sequences where each term after the first is obtained by adding a constant difference, focusing on their properties, nth term, and sum formulas.
Arithmetic Progressions - Practice Worksheet
Strengthen your foundation with key concepts and basic applications.
This worksheet covers essential long-answer questions to help you build confidence in Arithmetic Progressions from Mathematics for Class X (Mathematics).
Basic comprehension exercises
Strengthen your understanding with fundamental questions about the chapter.
Questions
Define Arithmetic Progression (AP) and explain its significance with a real-life example.
Think about situations where a fixed amount is added or subtracted regularly.
Solution
An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant difference, called the common difference, to the preceding term. This means if the first term is 'a' and the common difference is 'd', the sequence is a, a+d, a+2d, a+3d, and so on. AP is significant because it models many real-life situations where quantities change by a constant amount over time. For example, consider the salary of an employee who gets a fixed annual increment. If the starting salary is ₹20,000 and the annual increment is ₹2,000, the salaries over the years form an AP: 20,000, 22,000, 24,000, etc. This helps in predicting future salaries and planning finances. The general form of an AP is crucial for solving problems related to sequences and series in mathematics.
Find the 10th term of the AP: 3, 7, 11, 15, ...
Use the formula for the nth term of an AP: a_n = a + (n-1)d.
Solution
To find the 10th term of the given AP, we first identify the first term (a) and the common difference (d). Here, a = 3 and d = 7 - 3 = 4. The nth term of an AP is given by the formula: a_n = a + (n-1)d. Plugging in the values, a_10 = 3 + (10-1)*4 = 3 + 9*4 = 3 + 36 = 39. Therefore, the 10th term of the AP is 39. This formula is derived from the definition of AP, where each term increases by the common difference. Understanding this formula is essential for solving problems related to finding specific terms in an AP.
How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Use the sum formula for AP and solve the resulting quadratic equation.
Solution
To find the number of terms (n) that sum up to 636 in the given AP, we use the sum formula for an AP: S_n = n/2 [2a + (n-1)d]. Here, a = 9, d = 17 - 9 = 8, and S_n = 636. Plugging in the values, 636 = n/2 [2*9 + (n-1)*8] => 636 = n/2 [18 + 8n - 8] => 636 = n/2 [10 + 8n] => 636 = 5n + 4n^2. Rearranging, we get 4n^2 + 5n - 636 = 0. Solving this quadratic equation using the quadratic formula, n = [-5 ± √(25 + 10176)] / 8 = [-5 ± √10201] / 8 = [-5 ± 101] / 8. Taking the positive root, n = (96)/8 = 12. Therefore, 12 terms of the AP must be taken to get a sum of 636. This problem demonstrates the application of the sum formula and quadratic equations in AP.
The first term of an AP is 5, the last term is 45, and the sum is 400. Find the number of terms and the common difference.
First find the number of terms using the sum formula, then find the common difference using the nth term formula.
Solution
Given: first term (a) = 5, last term (l) = 45, sum (S_n) = 400. We use the sum formula for AP: S_n = n/2 (a + l). Plugging in the values, 400 = n/2 (5 + 45) => 400 = n/2 * 50 => 400 = 25n => n = 16. Now, to find the common difference (d), we use the nth term formula: l = a + (n-1)d. Plugging in the values, 45 = 5 + (16-1)d => 45 = 5 + 15d => 40 = 15d => d = 40/15 = 8/3. Therefore, the number of terms is 16, and the common difference is 8/3. This problem combines the use of sum and nth term formulas to find missing parameters in an AP.
Find the sum of the first 15 multiples of 8.
Identify the AP formed by the multiples and use the sum formula.
Solution
The first 15 multiples of 8 form an AP: 8, 16, 24, ..., up to 15 terms. Here, a = 8, d = 8, and n = 15. The sum of the first n terms of an AP is given by S_n = n/2 [2a + (n-1)d]. Plugging in the values, S_15 = 15/2 [2*8 + (15-1)*8] = 15/2 [16 + 112] = 15/2 * 128 = 15 * 64 = 960. Therefore, the sum of the first 15 multiples of 8 is 960. This problem illustrates how AP can be used to find the sum of multiples of a number, which is a common application in mathematics.
Which term of the AP: 21, 18, 15, ... is -81?
Use the nth term formula and solve for n.
Solution
To find which term of the given AP is -81, we use the nth term formula: a_n = a + (n-1)d. Here, a = 21, d = 18 - 21 = -3, and a_n = -81. Plugging in the values, -81 = 21 + (n-1)*(-3) => -81 = 21 - 3n + 3 => -81 = 24 - 3n => -105 = -3n => n = 35. Therefore, the 35th term of the AP is -81. This problem demonstrates how to find the position of a specific term in an AP using the nth term formula.
The sum of the first n terms of an AP is given by S_n = 4n - n^2. Find the first term and the common difference.
Find the first term by plugging n=1 into the sum formula, then find the second term and common difference.
Solution
Given the sum of the first n terms as S_n = 4n - n^2. The first term (a) is S_1 = 4*1 - 1^2 = 4 - 1 = 3. The sum of the first two terms is S_2 = 4*2 - 2^2 = 8 - 4 = 4. Therefore, the second term is S_2 - S_1 = 4 - 3 = 1. The common difference (d) is the second term minus the first term: d = 1 - 3 = -2. Thus, the first term is 3, and the common difference is -2. This problem shows how to derive the first term and common difference from the sum formula of an AP.
Find the sum of the odd numbers between 0 and 50.
Identify the AP formed by the odd numbers and use the sum formula.
Solution
The odd numbers between 0 and 50 form an AP: 1, 3, 5, ..., 49. Here, a = 1, l = 49. First, find the number of terms (n). Using the nth term formula: l = a + (n-1)d, where d = 2. So, 49 = 1 + (n-1)*2 => 49 = 1 + 2n - 2 => 49 = 2n - 1 => 50 = 2n => n = 25. Now, use the sum formula for AP: S_n = n/2 (a + l). Plugging in the values, S_25 = 25/2 (1 + 49) = 25/2 * 50 = 25 * 25 = 625. Therefore, the sum of the odd numbers between 0 and 50 is 625. This problem demonstrates the application of AP in summing a sequence of numbers with a common difference.
The 4th term of an AP is 0. Prove that the 25th term is triple the 11th term.
Express the 25th and 11th terms in terms of the common difference and compare them.
Solution
Given the 4th term (a_4) = 0. Using the nth term formula: a_n = a + (n-1)d. So, a_4 = a + 3d = 0 => a = -3d. Now, the 25th term (a_25) = a + 24d = -3d + 24d = 21d. The 11th term (a_11) = a + 10d = -3d + 10d = 7d. Therefore, a_25 = 21d = 3 * 7d = 3 * a_11. Hence, the 25th term is triple the 11th term. This problem shows how to use the nth term formula to establish relationships between different terms in an AP.
A man saves ₹100 in the first month, ₹150 in the second month, ₹200 in the third month, and so on. How much will he save in 2 years?
Identify the AP formed by the savings and use the sum formula for 24 terms.
Solution
The savings form an AP: 100, 150, 200, ... with a = 100, d = 50. In 2 years, there are 24 months, so n = 24. The sum of the first n terms of an AP is S_n = n/2 [2a + (n-1)d]. Plugging in the values, S_24 = 24/2 [2*100 + (24-1)*50] = 12 [200 + 1150] = 12 * 1350 = 16,200. Therefore, the man will save ₹16,200 in 2 years. This problem illustrates how AP can model savings over time, which is a practical application in personal finance.
Arithmetic Progressions - Mastery Worksheet
Advance your understanding through integrative and tricky questions.
This worksheet challenges you with deeper, multi-concept long-answer questions from Arithmetic Progressions to prepare for higher-weightage questions in Class X Mathematics.
Intermediate analysis exercises
Deepen your understanding with analytical questions about themes and characters.
Questions
Explain the concept of an Arithmetic Progression (AP) and how it differs from other sequences. Provide examples to illustrate your explanation.
Focus on the definition of AP and compare it with other sequences like geometric or harmonic progressions.
Solution
An Arithmetic Progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. This difference is known as the common difference (d). Unlike geometric sequences where terms are multiplied by a common ratio, AP terms are added by a common difference. Example: 2, 5, 8, 11, ... is an AP with a common difference of 3.
Find the 10th term of the AP: 3, 7, 11, 15, ... and explain the steps involved in the calculation.
Remember the formula for the nth term of an AP and ensure to substitute the correct values for a1, d, and n.
Solution
To find the 10th term (a10) of the AP, first identify the first term (a1) as 3 and the common difference (d) as 4 (7-3=4). Use the formula for the nth term of an AP: an = a1 + (n-1)d. Substituting the values: a10 = 3 + (10-1)*4 = 3 + 36 = 39.
Compare the sum of the first n terms of an AP with the sum of the first n terms of a geometric progression (GP). Use examples to highlight the differences.
Focus on the formulas and how the common difference in AP and common ratio in GP affect the sum.
Solution
The sum of the first n terms of an AP is given by Sn = n/2 [2a1 + (n-1)d], where a1 is the first term and d is the common difference. For a GP, the sum is Sn = a1(1 - r^n)/(1 - r), where r is the common ratio. Example: For AP 2, 5, 8, ... sum of first 3 terms is 15. For GP 2, 6, 18, ... sum of first 3 terms is 26.
A ladder has rungs 25 cm apart. The lengths of the rungs decrease uniformly from 45 cm at the bottom to 25 cm at the top. If the distance between the top and bottom rung is 2.5 m, how many rungs does the ladder have?
Consider the distance between the first and last rung and how the rungs are spaced.
Solution
First, convert the total height to cm: 2.5 m = 250 cm. The number of gaps between rungs is 250/25 = 10. Therefore, the number of rungs is 10 + 1 = 11.
The sum of the first n terms of an AP is given by Sn = 3n^2 + 5n. Find the 10th term of the AP.
Use the relationship between the sum of terms and individual terms: an = Sn - Sn-1.
Solution
First, find the sum of the first 9 terms (S9) and the first 10 terms (S10). S9 = 3*(9)^2 + 5*9 = 243 + 45 = 288. S10 = 3*(10)^2 + 5*10 = 300 + 50 = 350. The 10th term is S10 - S9 = 350 - 288 = 62.
If the 5th term of an AP is 17 and the 9th term is 33, find the sum of the first 20 terms.
First find the common difference and first term, then apply the sum formula.
Solution
Using the nth term formula: a5 = a1 + 4d = 17 and a9 = a1 + 8d = 33. Subtract the first equation from the second: 4d = 16 => d = 4. Substitute back to find a1: a1 + 16 = 17 => a1 = 1. Now, use the sum formula: S20 = 20/2 [2*1 + (20-1)*4] = 10 [2 + 76] = 10 * 78 = 780.
Explain why the sequence 1, 4, 9, 16, ... is not an AP. What type of sequence is it?
Check the difference between terms and identify the pattern of the sequence.
Solution
The sequence 1, 4, 9, 16, ... is not an AP because the difference between consecutive terms is not constant (4-1=3, 9-4=5, 16-9=7). It is a sequence of perfect squares, specifically a quadratic sequence where each term is n^2.
A contract specifies a penalty for delay of completion beyond a certain date: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc. How much penalty is paid for 30 days of delay?
Recognize the AP and use the sum formula with n=30.
Solution
This forms an AP with a1 = 200 and d = 50. The penalty for 30 days is the sum of the first 30 terms: S30 = 30/2 [2*200 + (30-1)*50] = 15 [400 + 1450] = 15 * 1850 = ₹27,750.
The sum of the first 15 terms of an AP is 300 and the sum of the next 15 terms is 600. Find the first term and common difference.
Set up equations for S15 and S30, then solve the system of equations.
Solution
Let S15 = 15/2 [2a1 + 14d] = 300 => 2a1 + 14d = 40. S30 = 30/2 [2a1 + 29d] = 300 + 600 = 900 => 2a1 + 29d = 60. Subtract the first equation from the second: 15d = 20 => d = 4/3. Substitute back to find a1: 2a1 + 14*(4/3) = 40 => 2a1 = 40 - 56/3 = 64/3 => a1 = 32/3.
In an AP, the sum of the first 10 terms is 150 and the sum of the next 10 terms is 550. Find the AP.
Use the sum formulas for S10 and S20 to find a1 and d.
Solution
S10 = 10/2 [2a1 + 9d] = 150 => 2a1 + 9d = 30. S20 = 20/2 [2a1 + 19d] = 150 + 550 = 700 => 2a1 + 19d = 70. Subtract the first equation from the second: 10d = 40 => d = 4. Substitute back to find a1: 2a1 + 36 = 30 => 2a1 = -6 => a1 = -3. The AP is -3, 1, 5, 9, ...
Arithmetic Progressions - Challenge Worksheet
Push your limits with complex, exam-level long-form questions.
The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Arithmetic Progressions in Class X Mathematics.
Advanced critical thinking
Test your mastery with complex questions that require critical analysis and reflection.
Questions
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2.5 m apart, what is the length of the wood required for the rungs?
Consider the number of rungs and the arithmetic progression formed by their lengths.
Solution
To find the total length of wood required, first determine the number of rungs. The distance between the top and bottom rungs is 2.5 m (250 cm), and the rungs are 25 cm apart. The number of rungs is (250 / 25) + 1 = 11. The lengths of the rungs form an AP with the first term (a) as 45 cm and the last term (l) as 25 cm. The sum of the lengths (S) of the rungs is given by S = n/2 (a + l) = 11/2 (45 + 25) = 11/2 * 70 = 385 cm. Therefore, the total length of wood required is 385 cm.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.
Express the third and seventh terms in terms of 'a' and 'd' and solve the given conditions.
Solution
Let the first term be 'a' and the common difference be 'd'. The third term (a3) is a + 2d, and the seventh term (a7) is a + 6d. Given that a3 + a7 = 6 and a3 * a7 = 8. Substituting the expressions for a3 and a7, we get (a + 2d) + (a + 6d) = 2a + 8d = 6, and (a + 2d)(a + 6d) = 8. Solving these equations simultaneously, we find a = 1 and d = 0.5. The sum of the first sixteen terms (S16) is S16 = 16/2 [2*1 + (16-1)*0.5] = 8 [2 + 7.5] = 8 * 9.5 = 76.
Which term of the AP: 121, 117, 113, ..., is its first negative term?
Find the term where the expression for the nth term becomes less than zero.
Solution
The first term (a) is 121, and the common difference (d) is -4. The nth term (an) of an AP is given by an = a + (n-1)d. We need to find the smallest n for which an < 0. So, 121 + (n-1)(-4) < 0 => 121 - 4(n-1) < 0 => 125 < 4n => n > 31.25. Since n must be an integer, the first negative term is the 32nd term.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
Set the sum of numbers before x equal to the sum after x and solve for x.
Solution
The sum of numbers from 1 to (x-1) is (x-1)x/2, and the sum from (x+1) to 49 is (49-x)(x+50)/2. Setting these equal: (x-1)x/2 = (49-x)(x+50)/2. Simplifying, we get x^2 - x = 49x + 2450 - x^2 - 50x => 2x^2 = 2450 => x^2 = 1225 => x = 35 (since x must be between 1 and 49).
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. Calculate the total volume of concrete required to build the terrace.
Calculate the volume of one step and multiply by the number of steps, considering if the dimensions change.
Solution
The volume of each step can be calculated as length * rise * tread. For the first step: 50 * 1/4 * 1/2 = 6.25 m³. The dimensions of the steps form an AP where the rise increases by 1/4 m for each subsequent step. However, since each step has the same dimensions, the total volume is 15 * 6.25 = 93.75 m³.
If the sum of the first n terms of an AP is given by Sn = 4n - n^2, find the first term and the common difference.
Use the given sum formula to find the first term and the difference between consecutive terms.
Solution
The sum of the first n terms (Sn) is 4n - n^2. The first term (a1) is S1 = 4*1 - 1^2 = 3. The sum of the first two terms (S2) is 4*2 - 2^2 = 4, so the second term (a2) is S2 - S1 = 4 - 3 = 1. The common difference (d) is a2 - a1 = 1 - 3 = -2.
The sum of the first 16 terms of an AP is 112 and the sum of the next 14 terms is 518. Find the AP.
Set up equations for the sums of the specified terms and solve for 'a' and 'd'.
Solution
Let the first term be 'a' and the common difference be 'd'. The sum of the first 16 terms (S16) is 16/2 [2a + 15d] = 112 => 8(2a + 15d) = 112 => 2a + 15d = 14. The sum of the next 14 terms (terms 17 to 30) is S30 - S16 = 518 => 30/2 [2a + 29d] - 112 = 518 => 15(2a + 29d) = 630 => 2a + 29d = 42. Solving these two equations, we find d = 2 and a = -8. The AP is -8, -6, -4, -2, ...
A man starts repaying a loan as a first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount will he pay in the 30th installment?
Recognize the installments as an AP and use the formula for the nth term.
Solution
The installments form an AP with the first term (a) as 100 and common difference (d) as 5. The 30th installment (a30) is a + (30-1)d = 100 + 29*5 = 100 + 145 = Rs. 245.
A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ..., up to 13 semicircles. What is the total length of such a spiral?
Calculate the circumference of each semicircle and sum them up as an AP.
Solution
The lengths of the semicircles form an AP with the first term (a) as π*0.5, and common difference (d) as π*0.5. The number of terms (n) is 13. The sum of the lengths (S) is S = n/2 [2a + (n-1)d] = 13/2 [π + 12*π*0.5] = 13/2 [π + 6π] = 13/2 * 7π = 45.5π cm. Using π ≈ 22/7, S ≈ 45.5 * 22/7 = 143 cm.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, and so on. What is the total distance the competitor has to run?
Consider the round trip for each potato and sum the distances as an AP.
Solution
The distances to the potatoes form an AP: 5 m, 8 m, 11 m, ..., up to the 10th potato. The nth term is 5 + 3(n-1). The total distance is twice the sum of these distances (since the competitor runs to the potato and back). The sum of the AP is 10/2 [2*5 + 9*3] = 5 [10 + 27] = 185 m. Total distance is 2 * 185 = 370 m.
Real Numbers encompass all rational and irrational numbers, forming a complete and continuous number line essential for various mathematical concepts.
Explore the world of Polynomials, understanding their types, degrees, and operations to solve algebraic expressions and equations effectively.
Explore the methods to solve a pair of linear equations in two variables, including graphical, substitution, elimination, and cross-multiplication techniques.
Explore the world of quadratic equations, learning to solve them using various methods like factoring, completing the square, and the quadratic formula.
