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Maharashtra SSC Class 10 Mathematics Chapter 3 Arithmetic Progression: textbook solutions

Solutions written by Shishya, checked, not the board's official answers

Step-by-step working for the practice-set questions of Chapter 3, “Arithmetic Progression”, in Mathematics Part I (Algebra), Standard X (Maharashtra SSC, English medium; the chapter is on pages 56-80 of the printed book). Each question is in our own short words and cited by its set and number; open the textbook for the exact wording and figures.

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How these were checked: drafted with AI on 28 Oct 2026. A solution is shown only if a calculator check recomputed every final value from the question's own numbers and every arithmetic step in the working agreed. 47 of 51 questions in this chapter passed; the others are listed below as not solved here. These are not the board's official answers: the textbook and your teacher are the authority, and if something here disagrees with the book, follow the book and tell us.

Textbook PDF read on 28 Oct 2026 from the publisher's official site.

Practice Set 3.1

Practice Set 3.1, Q1
Which of these sequences are APs? Give d for each AP: (1) 2, 4, 6, 8, … (2) 2, 5/2, 3, 7/2, … (3) −10, −6, −2, 2, … (4) 0.3, 0.33, 0.0333, … (5) 0, −4, −8, −12, … (6) −1/5, −1/5, −1/5, … (7) 3, 3 + √2, 3 + 2√2, 3 + 3√2, … (8) 127, 132, 137, …
1. (1) Check the difference between consecutive terms: 4 − 2 = 2, 6 − 4 = 2, 8 − 6 = 2.
2. (1) Since the common difference is constant, it is an AP with d = 2.
3. (2) Check the difference between consecutive terms: 5/2 − 2 = 1/2, 3 − 5/2 = 1/2, 7/2 − 3 = 1/2.
4. (2) Since the common difference is constant, it is an AP with d = 1/2.
5. (3) Check the difference between consecutive terms: −6 − (−10) = 4, −2 − (−6) = 4, 2 − (−2) = 4.
6. (3) Since the common difference is constant, it is an AP with d = 4.
7. (4) Check the difference between consecutive terms: 0.33 − 0.3 = 0.03, 0.0333 − 0.33 = −0.2967.
8. (4) Since the differences are not equal, it is not an AP.
9. (5) Check the difference between consecutive terms: −4 − 0 = −4, −8 − (−4) = −4, −12 − (−8) = −4.
10. (5) Since the common difference is constant, it is an AP with d = −4.
11. (6) Check the difference between consecutive terms: −1/5 − (−1/5) = 0, −1/5 − (−1/5) = 0.
12. (6) Since the common difference is constant, it is an AP with d = 0.
13. (7) Check the difference between consecutive terms: (3 + √2) − 3 = √2, (3 + 2√2) − (3 + √2) = √2, (3 + 3√2) − (3 + 2√2) = √2.
14. (7) Since the common difference is constant, it is an AP with d = √2.
15. (8) Check the difference between consecutive terms: 132 − 127 = 5, 137 − 132 = 5.
16. (8) Since the common difference is constant, it is an AP with d = 5.
Answer: (1) d = 2; (2) d = 1/2; (3) d = 4; (4) not an AP; (5) d = -4; (6) d = 0; (7) d = √2; (8) d = 5
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Practice Set 3.1, Q2
Write the first four terms of the AP with: (1) a = 10, d = 5 (2) a = −3, d = 0 (3) a = −7, d = 1/2 (4) a = −1.25, d = 3 (5) a = 6, d = −3 (6) a = −19, d = −4.
1. (1) First term a = 10.
2. (1) Second term = a + d = 10 + 5 = 15.
3. (1) Third term = a + 2d = 10 + 10 = 20.
4. (1) Fourth term = a + 3d = 10 + 15 = 25.
5. (2) First term a = -3.
6. (2) Second term = a + d = -3 + 0 = -3.
7. (2) Third term = a + 2d = -3 + 0 = -3.
8. (2) Fourth term = a + 3d = -3 + 0 = -3.
9. (3) First term a = -7.
10. (3) Second term = a + d = -7 + 0.5 = -6.5.
11. (3) Third term = a + 2d = -7 + 1 = -6.
12. (3) Fourth term = a + 3d = -7 + 1.5 = -5.5.
13. (4) First term a = -1.25.
14. (4) Second term = a + d = -1.25 + 3 = 1.75.
15. (4) Third term = a + 2d = -1.25 + 6 = 4.75.
16. (4) Fourth term = a + 3d = -1.25 + 9 = 7.75.
17. (5) First term a = 6.
18. (5) Second term = a + d = 6 - 3 = 3.
19. (5) Third term = a + 2d = 6 - 6 = 0.
20. (5) Fourth term = a + 3d = 6 - 9 = -3.
21. (6) First term a = -19.
22. (6) Second term = a + d = -19 - 4 = -23.
23. (6) Third term = a + 2d = -19 - 8 = -27.
24. (6) Fourth term = a + 3d = -19 - 12 = -31.
Answer: (1) 10, 15, 20, 25; (2) -3, -3, -3, -3; (3) -7, -6.5, -6, -5.5; (4) -1.25, 1.75, 4.75, 7.75; (5) 6, 3, 0, -3; (6) -19, -23, -27, -31
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Practice Set 3.1, Q3
Find the first term a and common difference d of: (1) 5, 1, −3, −7, … (2) 0.6, 0.9, 1.2, 1.5, … (3) 127, 135, 143, 151, … (4) 1/4, 3/4, 5/4, 7/4, …
1. (1) The first term is the first number in the sequence.
2. (1) a = 5
3. (1) The common difference is found by subtracting the first term from the second term.
4. (1) d = 1 − 5
5. (1) d = −4
6. (2) The first term is the first number in the sequence.
7. (2) a = 0.6
8. (2) The common difference is found by subtracting the first term from the second term.
9. (2) d = 0.9 − 0.6
10. (2) d = 0.3
11. (3) The first term is the first number in the sequence.
12. (3) a = 127
13. (3) The common difference is found by subtracting the first term from the second term.
14. (3) d = 135 − 127
15. (3) d = 8
16. (4) The first term is the first number in the sequence.
17. (4) a = 1/4
18. (4) The common difference is found by subtracting the first term from the second term.
19. (4) d = 3/4 − 1/4
20. (4) d = 2/4
21. (4) d = 1/2
Answer: (1) a = 5, d = -4; (2) a = 0.6, d = 0.3; (3) a = 127, d = 8; (4) a = 1/4, d = 1/2
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Practice Set 3.2

Practice Set 3.2, Q1
Read the terms and find d (and a) for: (i) 1, 8, 15, 22, … (ii) 3, 6, 9, 12, … (iii) −3, −8, −13, −18, … (iv) 70, 60, 50, 40, …
1. (i) The first term a is 1.
2. (i) The common difference d is 8 − 1 = 7.
3. (ii) The first term a is 3.
4. (ii) The common difference d is 6 − 3 = 3.
5. (iii) The first term a is −3.
6. (iii) The common difference d is −8 − (−3) = −5.
7. (iv) The first term a is 70.
8. (iv) The common difference d is 60 − 70 = −10.
Answer: (i) a = 1, d = 7; (ii) a = 3, d = 3; (iii) a = -3, d = -5; (iv) a = 70, d = -10
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Practice Set 3.2, Q2
Is −12, −5, 2, 9, 16, 23, 30, … an AP? If so, find its 20th term.
1. Check the common difference between consecutive terms.
2. d = −5 − (−12) = 7.
3. d = 2 − (−5) = 7.
4. d = 9 − 2 = 7.
5. Since the difference is constant, it is an AP.
6. Identify the first term a = −12 and common difference d = 7.
7. Use the formula for the nth term: aₙ = a + (n − 1)d.
8. Substitute n = 20, a = −12, and d = 7 into the formula.
9. a₂₀ = −12 + (20 − 1) × 7.
10. a₂₀ = −12 + 19 × 7.
11. a₂₀ = −12 + 133.
12. a₂₀ = 121.
Answer: Yes, it is an AP; its 20th term is 121
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Practice Set 3.2, Q3
For the AP 12, 16, 20, 24, …, find the 24th term.
1. The first term of the AP is a = 12.
2. The common difference is d = 16 − 12 = 4.
3. The formula for the nth term is aₙ = a + (n − 1)d.
4. Substitute n = 24, a = 12, and d = 4 into the formula.
5. a₂₄ = 12 + (24 − 1) × 4.
6. a₂₄ = 12 + 23 × 4.
7. a₂₄ = 12 + 92.
8. a₂₄ = 104.
Answer: 104
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Practice Set 3.2, Q4
Find the 19th term of the AP 7, 13, 19, 25, …
1. Given AP: 7, 13, 19, 25, ...
2. First term a = 7
3. Common difference d = 13 − 7 = 6
4. We need the 19th term, so n = 19
5. Formula: aₙ = a + (n − 1)d
6. a₁₉ = 7 + (19 − 1) × 6
7. a₁₉ = 7 + 18 × 6
8. a₁₉ = 7 + 108
9. a₁₉ = 115
Answer: 19th term = 115
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Practice Set 3.2, Q5
Find the 27th term of the AP 9, 4, −1, −6, −11, …
1. Given AP: 9, 4, −1, −6, −11, …
2. First term a = 9
3. Common difference d = 4 − 9 = −5
4. We need the 27th term, so n = 27
5. Formula: aₙ = a + (n − 1)d
6. a₂₇ = 9 + (27 − 1)(−5)
7. a₂₇ = 9 + 26 × (−5)
8. a₂₇ = 9 − 130
9. a₂₇ = −121
Answer: The 27th term is −121
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Practice Set 3.2, Q6
How many three-digit natural numbers are divisible by 5?
1. The smallest three-digit number divisible by 5 is 100.
2. The largest three-digit number divisible by 5 is 995.
3. This forms an arithmetic progression with first term a = 100, common difference d = 5, and last term l = 995.
4. Using the formula for the nth term, l = a + (n − 1)d.
5. Substitute the values: 995 = 100 + (n − 1) × 5.
6. Subtract 100 from both sides: 895 = (n − 1) × 5.
7. Divide by 5: 179 = n − 1.
8. Add 1 to both sides: n = 180.
Answer: 180
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Practice Set 3.2, Q7
The 11th term of an AP is 16 and the 21st term is 29. Find the 41st term.
1. Given a₁₁ = 16 and a₂₁ = 29.
2. Using the formula aₙ = a + (n−1)d, we have a + 10d = 16 and a + 20d = 29.
3. Subtracting the first equation from the second gives 10d = 13.
4. Solving for d gives d = 13/10.
5. Substituting d into a + 10d = 16 gives a + 13 = 16.
6. Solving for a gives a = 3.
7. The 41st term is a₄₁ = a + 40d.
8. Substituting the values gives a₄₁ = 3 + 40 × (13/10).
9. Calculating the product gives 40 × (13/10) = 52.
10. Adding the first term gives a₄₁ = 3 + 52 = 55.
Answer: 55
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Practice Set 3.2, Q8
In the AP 11, 8, 5, 2, …, which term is −151?
1. The first term a is 11.
2. The common difference d is 8 − 11 = −3.
3. The nth term formula is aₙ = a + (n − 1)d.
4. Substitute the values: −151 = 11 + (n − 1)(−3).
5. Subtract 11 from both sides: −151 − 11 = (n − 1)(−3).
6. Simplify the left side: −162 = (n − 1)(−3).
7. Divide both sides by −3: −162 ÷ (−3) = n − 1.
8. Calculate the division: 54 = n − 1.
9. Add 1 to both sides: n = 54 + 1.
10. Solve for n: n = 55.
Answer: 55
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Practice Set 3.2, Q9
How many natural numbers from 10 to 250 are divisible by 4?
1. The natural numbers divisible by 4 from 10 to 250 form an arithmetic progression.
2. The first term a is 12, as it is the smallest multiple of 4 greater than or equal to 10.
3. The common difference d is 4.
4. The last term l is 248, as it is the largest multiple of 4 less than or equal to 250.
5. Using the formula for the nth term of an AP: l = a + (n - 1)d.
6. Substitute the values: 248 = 12 + (n - 1)4.
7. Simplify the equation: 248 = 12 + 4n - 4.
8. Combine constants: 248 = 8 + 4n.
9. Subtract 8 from both sides: 240 = 4n.
10. Divide by 4: n = 60.
Answer: 60
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Practice Set 3.2, Q10
The 17th term of an AP is 7 more than its 10th term. Find the common difference.
1. Let the first term be a and the common difference be d.
2. The 17th term is a + 16d.
3. The 10th term is a + 9d.
4. According to the question, a + 16d = (a + 9d) + 7.
5. Subtract a from both sides: 16d = 9d + 7.
6. Subtract 9d from both sides: 7d = 7.
7. Divide by 7: d = 1.
Answer: The common difference is 1.
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Practice Set 3.3

Not solved here yet: Q9 (see the textbook).

Practice Set 3.3, Q1
An AP has first term 6 and common difference 3. Find S27.
1. Given: a = 6, d = 3, n = 27
2. Formula: S_n = n/2 [2a + (n-1)d]
3. S_27 = 27/2 [2(6) + (27-1)(3)]
4. S_27 = 27/2 [12 + 26(3)]
5. S_27 = 27/2 [12 + 78]
6. S_27 = 27/2 [90]
7. S_27 = 27 × 45
8. S_27 = 1215
Answer: S27 = 1215
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Practice Set 3.3, Q2
Find the sum of the first 123 even natural numbers.
1. The first 123 even natural numbers form an arithmetic progression: 2, 4, 6, ..., 246.
2. Here, the first term a = 2, the common difference d = 2, and the number of terms n = 123.
3. The sum of an arithmetic progression is given by S = n/2 × [2a + (n − 1)d].
4. Substitute the values: S = 123/2 × [2(2) + (123 − 1)(2)].
5. Simplify the term inside the brackets: 2(2) = 4 and (123 − 1)(2) = 122 × 2 = 244.
6. Add the terms inside the brackets: 4 + 244 = 248.
7. Now calculate S = 123/2 × 248.
8. Simplify the multiplication: 248 ÷ 2 = 124, so S = 123 × 124.
9. Compute the product: 123 × 124 = 15252.
Answer: 15252
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Practice Set 3.3, Q3
Find the sum of all even numbers between 1 and 350.
1. The even numbers between 1 and 350 are 2, 4, 6, ..., 348.
2. This forms an arithmetic progression with first term a = 2 and common difference d = 2.
3. The last term l = 348.
4. Using the formula l = a + (n − 1)d, we have 348 = 2 + (n − 1)2.
5. 348 − 2 = (n − 1)2
6. 346 = (n − 1)2
7. n − 1 = 346 ÷ 2
8. n − 1 = 173
9. n = 174
10. Sum S = n/2 × (a + l)
11. S = 174/2 × (2 + 348)
12. S = 87 × 350
13. S = 30450
Answer: 30450
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Practice Set 3.3, Q4
In an AP the 19th term is 52 and the 38th term is 128. Find the sum of the first 56 terms.
1. Given a_19 = 52 and a_38 = 128.
2. Using the formula a_n = a + (n-1)d, we have a + 18d = 52.
3. Similarly, a + 37d = 128.
4. Subtracting the first equation from the second gives 19d = 76.
5. Solving for d gives d = 4.
6. Substituting d = 4 into a + 18d = 52 gives a + 72 = 52.
7. Solving for a gives a = -20.
8. The sum of the first 56 terms is S_56 = 56/2 [2(-20) + (56-1)4].
9. Simplifying the expression gives S_56 = 28 [-40 + 220].
10. Calculating the final value gives S_56 = 28 × 180.
11. S_56 = 5040.
Answer: 5040
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Practice Set 3.3, Q5
Find the sum of the natural numbers between 1 and 140 that are divisible by 4.
1. The natural numbers between 1 and 140 divisible by 4 form an AP: 4, 8, 12, ..., 136.
2. Here, first term a = 4 and common difference d = 4.
3. The last term l is 136 because 140 is not divisible by 4 and 136 is the largest multiple of 4 less than 140.
4. Using the formula for the nth term, l = a + (n - 1)d, we substitute the values: 136 = 4 + (n - 1)4.
5. Simplifying the equation gives 132 = 4(n - 1).
6. Dividing by 4 yields n - 1 = 33, so n = 34.
7. The sum of the AP is given by S = n/2 × (a + l).
8. Substituting the values, S = 34/2 × (4 + 136).
9. S = 17 × 140.
10. S = 2380.
Answer: 2380
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Practice Set 3.3, Q6
The sum of the first 55 terms of an AP is 3300. Find its 28th term.
1. Given S₅₅ = 3300 and n = 55.
2. Using the formula Sₙ = n/2 [2a + (n−1)d].
3. 3300 = 55/2 [2a + 54d].
4. Dividing both sides by 55 gives 60 = 1/2 [2a + 54d].
5. Multiplying by 2 gives 120 = 2a + 54d.
6. Dividing by 2 gives 60 = a + 27d.
7. The 28th term is T₂₈ = a + (28−1)d = a + 27d.
8. Therefore, T₂₈ = 60.
Answer: 60
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Practice Set 3.3, Q7
Three consecutive terms of an AP add up to 27 and their product is 504. Find the terms (take them as a − d, a, a + d).
1. Let the three consecutive terms of the AP be a − d, a, and a + d.
2. The sum of the terms is given as 27, so (a − d) + a + (a + d) = 27.
3. Simplifying the sum gives 3a = 27, which means a = 9.
4. The product of the terms is given as 504, so (a − d) × a × (a + d) = 504.
5. Substitute a = 9 into the product equation to get (9 − d) × 9 × (9 + d) = 504.
6. Divide both sides by 9 to obtain (9 − d)(9 + d) = 56.
7. Expand the left side using the difference of squares formula: 81 − d² = 56.
8. Rearrange the equation to find d² = 81 − 56 = 25.
9. d² = 25, so d = 5 or d = −5; both give the same three terms (in opposite order), so take d = 5.
10. The first term is a − d = 9 − 5 = 4.
11. The second term is a = 9.
12. The third term is a + d = 9 + 5 = 14.
Answer: 4, 9, 14
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Practice Set 3.3, Q8
Four consecutive terms of an AP add up to 12, and the 3rd and 4th of them add up to 14. Find the terms (take them as a − d, a, a + d, a + 2d).
1. Let the four consecutive terms be a − d, a, a + d, and a + 2d.
2. Sum of the four terms is 12, so (a − d) + a + (a + d) + (a + 2d) = 12.
3. Simplifying the equation gives 4a + 2d = 12.
4. Dividing by 2 gives 2a + d = 6.
5. Sum of the 3rd and 4th terms is 14, so (a + d) + (a + 2d) = 14.
6. Simplifying the equation gives 2a + 3d = 14.
7. Subtracting the equation 2a + d = 6 from 2a + 3d = 14 gives 2d = 8.
8. Solving for d gives d = 4.
9. Substituting d = 4 into 2a + d = 6 gives 2a + 4 = 6.
10. Solving for a gives 2a = 2, so a = 1.
11. The first term is a − d = 1 − 4 = −3.
12. The second term is a = 1.
13. The third term is a + d = 1 + 4 = 5.
14. The fourth term is a + 2d = 1 + 8 = 9.
Answer: -3, 1, 5, 9
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Practice Set 3.4

Practice Set 3.4, Q1
Sanika saves ₹10 on 1 January 2016, ₹11 on the second day, ₹12 on the third day and so on. How much has she saved in all by 31 December 2016?
1. The year 2016 is a leap year, so it has 366 days.
2. The savings form an arithmetic progression with first term a = 10 and common difference d = 1.
3. The number of terms is n = 366.
4. The sum of an arithmetic progression is given by S = n/2 [2a + (n − 1)d].
5. Substitute the values: S = 366/2 [2(10) + (366 − 1)(1)].
6. Simplify the bracket: 2(10) + 365(1) = 20 + 365 = 385.
7. Multiply: S = 183 × 385.
8. Calculate the product: 183 × 385 = 70455.
Answer: Sanika has saved ₹70455 in all by 31 December 2016.
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Practice Set 3.4, Q2
A man borrows ₹8000 and repays it with a total interest of ₹1360 in 12 monthly instalments, each instalment ₹40 less than the one before. Find the first and the last instalment.
1. Let the first instalment be x rupees.
2. The instalments form an arithmetic progression with common difference d = -40.
3. The sum of 12 instalments is 8000 + 1360 = 9360.
4. Using the sum formula S = n/2 [2a + (n-1)d], we have 9360 = 12/2 [2x + 11(-40)].
5. Simplifying gives 9360 = 6 [2x - 440].
6. Dividing by 6 gives 1560 = 2x - 440.
7. Adding 440 to both sides gives 2x = 2000.
8. Dividing by 2 gives x = 1000.
9. The last instalment is the 12th term: a_12 = x + 11d.
10. Substituting values gives a_12 = 1000 + 11(-40).
11. Calculating gives a_12 = 1000 - 440 = 560.
Answer: First instalment ₹1000; last instalment ₹560
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Practice Set 3.4, Q3
Sachin invests ₹5000 in the first year, ₹7000 in the second year, ₹9000 in the third year and so on. How much does he invest in 12 years?
1. The investments form an arithmetic progression with first term a = 5000 and common difference d = 7000 − 5000 = 2000
2. We need the total investment over 12 years, so n = 12
3. Use the sum formula Sₙ = n/2 × [2a + (n − 1)d]
4. S₁₂ = 12/2 × [2 × 5000 + (12 − 1) × 2000]
5. S₁₂ = 6 × [10000 + 11 × 2000]
6. S₁₂ = 6 × [10000 + 22000]
7. S₁₂ = 6 × 32000
8. S₁₂ = 192000
Answer: Sachin invests ₹192000 in 12 years
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Practice Set 3.4, Q4
An auditorium has 27 rows: 20 seats in the first row, 22 in the second, 24 in the third and so on. Find the number of seats in the 15th row and the total number of seats.
1. The number of seats in the rows forms an arithmetic progression with first term a = 20 and common difference d = 2.
2. To find the number of seats in the 15th row, use the formula for the nth term: a_n = a + (n - 1)d.
3. Substitute n = 15, a = 20, and d = 2 into the formula: a_15 = 20 + (15 - 1) × 2.
4. Calculate the term: a_15 = 20 + 14 × 2 = 20 + 28 = 48.
5. To find the total number of seats in 27 rows, use the sum formula: S_n = n/2 [2a + (n - 1)d].
6. Substitute n = 27, a = 20, and d = 2 into the formula: S_27 = 27/2 [2 × 20 + (27 - 1) × 2].
7. Simplify the expression inside the brackets: S_27 = 27/2 [40 + 26 × 2] = 27/2 [40 + 52].
8. Add the values inside the brackets: S_27 = 27/2 [92].
9. Calculate the final sum: S_27 = 27 × 46 = 1242.
Answer: 48 seats in the 15th row; 1242 seats in all
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Practice Set 3.4, Q5
Temperatures from Monday to Saturday are in AP. Monday's plus Saturday's temperature is 5 °C more than Tuesday's plus Saturday's. Wednesday's temperature is −30 °C. Find the temperatures of the other five days.
1. Let the temperatures from Monday to Saturday be a, a+d, a+2d, a+3d, a+4d, a+5d.
2. Monday's plus Saturday's temperature is 5 °C more than Tuesday's plus Saturday's; Saturday's temperature is on both sides, so Monday's temperature is 5 °C more than Tuesday's.
3. a = (a + d) + 5
4. a = a + d + 5
5. d = -5
6. Given that Wednesday's temperature is -30 °C.
7. a + 2d = -30
8. a + 2(-5) = -30
9. a - 10 = -30
10. a = -20
11. Monday's temperature = a = -20 °C
12. Tuesday's temperature = a + d = -20 + (-5) = -25 °C
13. Thursday's temperature = a + 3d = -20 + 3(-5) = -35 °C
14. Friday's temperature = a + 4d = -20 + 4(-5) = -40 °C
15. Saturday's temperature = a + 5d = -20 + 5(-5) = -45 °C
Answer: Monday -20 °C, Tuesday -25 °C, Thursday -35 °C, Friday -40 °C, Saturday -45 °C
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Practice Set 3.4, Q6
Trees are planted in a triangular plot: 1 tree in the first row, 2 in the second, 3 in the third and so on. How many trees are there in 25 rows?
1. The number of trees in each row forms an arithmetic progression: 1, 2, 3, ...
2. The first term a is 1 and the common difference d is 1.
3. We need to find the sum of the first 25 terms, so n = 25.
4. The formula for the sum of an AP is S_n = n/2 [2a + (n-1)d].
5. Substitute the values into the formula: S_25 = 25/2 [2(1) + (25-1)(1)].
6. Simplify the expression inside the brackets: 2 + 24 = 26.
7. Calculate the product: S_25 = 25/2 × 26.
8. Simplify the fraction: 26 ÷ 2 = 13.
9. Calculate the final sum: 25 × 13 = 325.
Answer: 325
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Problem Set 3

Not solved here yet: Q11, Q12, Q13 (see the textbook).

Problem Set 3, Q1(1)
Choose the correct option: the sequence −10, −6, −2, 2, … (A) is an AP, d = −16 (B) is an AP, d = 4 (C) is an AP, d = −4 (D) is not an AP.
1. The given sequence is −10, −6, −2, 2, …
2. Calculate the difference between the second and first terms: −6 − (−10) = 4
3. Calculate the difference between the third and second terms: −2 − (−6) = 4
4. Calculate the difference between the fourth and third terms: 2 − (−2) = 4
5. Since the common difference is constant, the sequence is an AP with d = 4
6. This matches option (B)
Answer: (B)
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Problem Set 3, Q1(2)
Choose the correct option: the first four terms of the AP with a = −2 and d = −2 are (A) −2, 0, 2, 4 (B) −2, 4, −8, 16 (C) −2, −4, −6, −8 (D) −2, −4, −8, −16.
1. The first term of the AP is a = −2.
2. The common difference is d = −2.
3. The second term is a + d = −2 + (−2) = −4.
4. The third term is a + 2d = −2 + 2(−2) = −6.
5. The fourth term is a + 3d = −2 + 3(−2) = −8.
6. The first four terms are −2, −4, −6, −8.
7. This matches option (C).
Answer: (C)
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Problem Set 3, Q1(3)
Choose the correct option: the sum of the first 30 natural numbers is (A) 464 (B) 465 (C) 462 (D) 461.
1. The sum of the first n natural numbers is given by the formula S = n(n + 1) ÷ 2.
2. Here, n = 30.
3. S = 30(30 + 1) ÷ 2
4. S = 30 × 31 ÷ 2
5. S = 930 ÷ 2
6. S = 465
7. This matches option (B).
Answer: 465 (B)
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Problem Set 3, Q1(4)
Choose the correct option: in an AP t7 = 4 and d = −4, so a = (A) 6 (B) 7 (C) 20 (D) 28.
1. Given t₇ = 4 and d = −4.
2. Using the formula tₙ = a + (n − 1)d.
3. Substitute n = 7, t₇ = 4, and d = −4: 4 = a + (7 − 1)(−4).
4. Simplify the term (7 − 1)(−4) to −24.
5. The equation becomes 4 = a − 24.
6. Add 24 to both sides: a = 4 + 24.
7. Calculate a = 28.
8. This matches option (D).
Answer: a = 28 (D)
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Problem Set 3, Q1(5)
Choose the correct option: in an AP a = 3.5, d = 0 and n = 101, so tn = (A) 0 (B) 3.5 (C) 103.5 (D) 104.5.
1. The formula for the nth term of an AP is tn = a + (n − 1)d.
2. Substitute the given values a = 3.5, d = 0, and n = 101 into the formula.
3. t101 = 3.5 + (101 − 1) × 0.
4. t101 = 3.5 + 100 × 0.
5. t101 = 3.5 + 0.
6. t101 = 3.5.
7. This matches option (B).
Answer: (B)
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Problem Set 3, Q1(6)
Choose the correct option: the first two terms of an AP are −3 and 4, so the 21st term is (A) −143 (B) 143 (C) 137 (D) 17.
1. Given first term a = −3 and second term a₂ = 4.
2. Common difference d = a₂ − a = 4 − (−3) = 7.
3. Formula for the nth term: aₙ = a + (n − 1)d.
4. Substitute n = 21, a = −3, d = 7: a₂₁ = −3 + (21 − 1) × 7.
5. a₂₁ = −3 + 20 × 7.
6. a₂₁ = −3 + 140.
7. a₂₁ = 137.
8. The 21st term is 137, which corresponds to option (C).
Answer: (C) 137
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Problem Set 3, Q1(7)
Choose the correct option: in an AP with d = 5, t18 − t13 = (A) 5 (B) 20 (C) 25 (D) 30.
1. The difference between the 18th and 13th terms is t18 − t13.
2. Using the formula tn = a + (n−1)d, we have t18 = a + 17d and t13 = a + 12d.
3. Subtracting these gives t18 − t13 = (a + 17d) − (a + 12d).
4. Simplifying the expression yields 5d.
5. Given d = 5, substitute to get 5 × 5.
6. Calculate 5 × 5 = 25.
7. This matches option (C).
Answer: (C) 25
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Problem Set 3, Q1(8)
Choose the correct option: the sum of the first five multiples of 3 is (A) 45 (B) 55 (C) 15 (D) 75.
1. The first five multiples of 3 are 3, 6, 9, 12, and 15.
2. Sum = 3 + 6 + 9 + 12 + 15
3. Sum = 45
4. This matches option (A)
Answer: (A) 45
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Problem Set 3, Q1(9)
Choose the correct option: the sum of the first 10 terms of 15, 10, 5, … is (A) −75 (B) −125 (C) 75 (D) 125.
1. The first term a is 15 and the common difference d is 10 − 15 = −5.
2. The number of terms n is 10.
3. Use the formula for the sum of an arithmetic progression: Sₙ = n/2 × [2a + (n − 1)d].
4. Substitute the values: S₁₀ = 10/2 × [2(15) + (10 − 1)(−5)].
5. Simplify the expression inside the brackets: 2(15) = 30 and (9)(−5) = −45.
6. Calculate the sum inside the brackets: 30 + (−45) = −15.
7. Multiply by n/2: 5 × (−15) = −75.
8. The sum is −75, which corresponds to option (A).
Answer: (A) −75
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Problem Set 3, Q1(10)
Choose the correct option: an AP has first term 1, last term 20 and sum 399, so n = (A) 42 (B) 38 (C) 21 (D) 19.
1. Given first term a = 1, last term l = 20, and sum S = 399.
2. The formula for the sum of an AP is S = n/2 × (a + l).
3. Substituting the values gives 399 = n/2 × (1 + 20).
4. Simplifying the bracket gives 399 = n/2 × 21.
5. Multiplying both sides by 2 gives 798 = 21n.
6. Dividing by 21 gives n = 798 ÷ 21.
7. Calculating the division gives n = 38.
8. This matches option (B).
Answer: n = 38 (B)
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Problem Set 3, Q2
Find the fourth term from the end of the AP −11, −8, −5, …, 49.
1. The given AP is −11, −8, −5, …, 49.
2. First term a = −11 and common difference d = −8 − (−11) = 3.
3. Last term l = 49.
4. The k-th term from the end of an AP is given by l − (k − 1)d.
5. For the 4th term from the end, k = 4.
6. 4th term from the end = 49 − (4 − 1) × 3.
7. 4th term from the end = 49 − 3 × 3.
8. 4th term from the end = 49 − 9.
9. 4th term from the end = 40.
Answer: 40
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Problem Set 3, Q3
In an AP the 10th term is 46 and the 5th and 7th terms add up to 52. Find the AP.
1. Let the first term be a and the common difference be d.
2. The 10th term is 46, so a + 9d = 46.
3. The sum of the 5th and 7th terms is 52, so (a + 4d) + (a + 6d) = 52.
4. Simplifying gives 2a + 10d = 52.
5. Dividing by 2 gives a + 5d = 26.
6. Subtracting a + 5d = 26 from a + 9d = 46 gives 4d = 20.
7. Solving for d gives d = 5.
8. Substituting d = 5 into a + 5d = 26 gives a + 25 = 26.
9. Solving for a gives a = 1.
10. The 1st term is a = 1.
11. The 2nd term is a + d = 1 + 5 = 6.
12. The 3rd term is a + 2d = 1 + 10 = 11.
13. The 4th term is a + 3d = 1 + 15 = 16.
Answer: 1, 6, 11, 16
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Problem Set 3, Q4
In an AP the 4th term is −15 and the 9th term is −30. Find the sum of the first 10 terms.
1. Given a₄ = −15 and a₉ = −30.
2. Using the formula aₙ = a + (n−1)d, we have a + 3d = −15 and a + 8d = −30.
3. Subtracting the first equation from the second gives 5d = −15.
4. Solving for d gives d = −3.
5. Substituting d = −3 into a + 3d = −15 gives a + 3(−3) = −15.
6. Simplifying gives a − 9 = −15, so a = −6.
7. The sum of the first 10 terms is S₁₀ = 10/2 [2a + (10−1)d].
8. Substituting the values gives S₁₀ = 5 [2(−6) + 9(−3)].
9. Calculating inside the brackets gives S₁₀ = 5 [−12 − 27].
10. Simplifying further gives S₁₀ = 5 [−39].
11. The final sum is S₁₀ = −195.
Answer: −195
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Problem Set 3, Q5
The APs 9, 7, 5, … and 24, 21, 18, … have equal nth terms. Find n and that term.
1. For the first AP, a = 9 and d = 7 − 9 = −2.
2. The nth term is aₙ = 9 + (n − 1)(−2).
3. Simplifying, aₙ = 9 − 2n + 2 = 11 − 2n.
4. For the second AP, a = 24 and d = 21 − 24 = −3.
5. The nth term is aₙ = 24 + (n − 1)(−3).
6. Simplifying, aₙ = 24 − 3n + 3 = 27 − 3n.
7. Since the nth terms are equal, 11 − 2n = 27 − 3n.
8. Adding 3n to both sides gives 11 + n = 27.
9. Subtracting 11 from both sides gives n = 16.
10. Substituting n = 16 into the first AP formula: a₁₆ = 11 − 2(16).
11. Calculating the value: a₁₆ = 11 − 32 = −21.
Answer: n = 16; the term is −21
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Problem Set 3, Q6
In an AP the 3rd and 8th terms add up to 7, and the 7th and 14th terms add up to −3. Find the 10th term.
1. Let the first term be a and the common difference be d.
2. The 3rd term is a + 2d and the 8th term is a + 7d.
3. Given their sum is 7, so (a + 2d) + (a + 7d) = 7.
4. Simplifying gives 2a + 9d = 7.
5. The 7th term is a + 6d and the 14th term is a + 13d.
6. Given their sum is -3, so (a + 6d) + (a + 13d) = -3.
7. Simplifying gives 2a + 19d = -3.
8. Subtract the first equation from the second: (2a + 19d) - (2a + 9d) = -3 - 7.
9. This gives 10d = -10.
10. Solving for d gives d = -1.
11. Substitute d = -1 into 2a + 9d = 7 to get 2a + 9(-1) = 7.
12. Solving gives 2a - 9 = 7, so 2a = 16 and a = 8.
13. The 10th term is given by a + 9d.
14. Substitute the values: 8 + 9(-1) = 8 - 9 = -1.
Answer: The 10th term is -1
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Problem Set 3, Q7
An AP has first term −5, last term 45 and sum 120. How many terms are there, and what is the common difference?
1. Given: a = −5, l = 45, S = 120.
2. Formula for sum of AP: S = (n/2)(a + l).
3. Substitute values: 120 = (n/2)(−5 + 45).
4. Simplify: 120 = (n/2)(40).
5. Simplify: 120 = 20n.
6. Solve for n: n = 120 ÷ 20.
7. n = 6.
8. Formula for last term: l = a + (n − 1)d.
9. Substitute values: 45 = −5 + (6 − 1)d.
10. Simplify: 45 = −5 + 5d.
11. Add 5 to both sides: 50 = 5d.
12. Solve for d: d = 50 ÷ 5.
13. d = 10.
Answer: 6 terms; d = 10
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Problem Set 3, Q8
The sum of the natural numbers from 1 to n is 36. Find n.
1. The sum of the first n natural numbers is given by the formula S = n(n + 1)/2.
2. Substitute the given sum 36 into the formula: 36 = n(n + 1)/2.
3. Multiply both sides by 2 to clear the denominator: 72 = n(n + 1).
4. Expand the right side to form a quadratic equation: n² + n = 72.
5. Rearrange the equation to standard form: n² + n − 72 = 0.
6. Factorize the quadratic expression: (n + 9)(n − 8) = 0.
7. Solve for n: n = −9 or n = 8.
8. Since n represents the count of natural numbers, it must be positive, so n = 8.
Answer: n = 8
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Problem Set 3, Q9
Divide 207 into three parts in AP such that the product of the two smaller parts is 4623.
1. Let the three parts in AP be a − d, a, and a + d.
2. Their sum is (a − d) + a + (a + d) = 3a.
3. Given the sum is 207, so 3a = 207.
4. Solving for a gives a = 207 ÷ 3 = 69.
5. The product of the two smaller parts is (a − d) × a = 4623.
6. Substitute a = 69 to get (69 − d) × 69 = 4623.
7. Divide both sides by 69 to get 69 − d = 4623 ÷ 69.
8. Calculate 4623 ÷ 69 = 67.
9. So 69 − d = 67, which gives d = 69 − 67 = 2.
10. The three parts are a − d = 69 − 2 = 67, a = 69, and a + d = 69 + 2 = 71.
Answer: 67, 69, 71
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Problem Set 3, Q10
An AP has 37 terms. The three middle terms add up to 225 and the last three terms add up to 429. Write the AP.
1. The three middle terms of an AP with 37 terms are the 18th, 19th, and 20th terms.
2. Their sum is 3 times the 19th term, so 3a_19 = 225.
3. a_19 = 225 ÷ 3 = 75.
4. The last three terms are the 35th, 36th, and 37th terms.
5. Their sum is 3 times the 36th term, so 3a_36 = 429.
6. a_36 = 429 ÷ 3 = 143.
7. Using the formula a_n = a + (n-1)d, we have a + 18d = 75 and a + 35d = 143.
8. Subtracting the first equation from the second gives 17d = 68.
9. d = 68 ÷ 17 = 4.
10. Substituting d = 4 into a + 18d = 75 gives a + 72 = 75.
11. a = 75 − 72 = 3.
12. The first three terms are 3, 3 + 4, and 3 + 8, which are 3, 7, and 11.
13. The 37th term is a + 36d = 3 + 144 = 147.
Answer: The AP is 3, 7, 11, …, 147
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Problem Set 3, Q14
₹1000 is invested at 10% simple interest. Is the total interest at the end of each year an AP? If so, find the interest after 20 years.
1. Principal P = 1000, Rate R = 10%, Time T = 1 year
2. Simple Interest for 1 year = P × R × T ÷ 100
3. SI = 1000 × 10 × 1 ÷ 100
4. SI = 100
5. Interest at end of 1st year = 100
6. Interest at end of 2nd year = 100 + 100 = 200
7. Interest at end of 3rd year = 200 + 100 = 300
8. The sequence of total interest is 100, 200, 300, ...
9. Common difference d = 200 − 100 = 100
10. Since the common difference is constant, it is an AP
11. To find interest after 20 years, use formula a_n = a + (n − 1)d
12. Here a = 100, n = 20, d = 100
13. a_20 = 100 + (20 − 1) × 100
14. a_20 = 100 + 19 × 100
15. a_20 = 100 + 1900
16. a_20 = 2000
Answer: Yes, the total interest is an AP; the interest after 20 years is ₹2000
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Other chapters solved for this book

Chapter 1: Linear Equations in Two Variables — 58 questions solved
Chapter 2: Quadratic Equations — 73 questions solved
Chapter 5: Probability — 25 questions solved

← All state-board textbook solutions

Maharashtra SSC Class 10 Mathematics Chapter 3 Arithmetic Progression: textbook solutions

Practice Set 3.1

Practice Set 3.1, Q1

Practice Set 3.1, Q2

Practice Set 3.1, Q3

Practice Set 3.2

Practice Set 3.2, Q1

Practice Set 3.2, Q2

Practice Set 3.2, Q3

Practice Set 3.2, Q4

Practice Set 3.2, Q5

Practice Set 3.2, Q6

Practice Set 3.2, Q7

Practice Set 3.2, Q8

Practice Set 3.2, Q9

Practice Set 3.2, Q10

Practice Set 3.3

Practice Set 3.3, Q1

Practice Set 3.3, Q2

Practice Set 3.3, Q3

Practice Set 3.3, Q4

Practice Set 3.3, Q5

Practice Set 3.3, Q6

Practice Set 3.3, Q7

Practice Set 3.3, Q8

Practice Set 3.4

Practice Set 3.4, Q1

Practice Set 3.4, Q2

Practice Set 3.4, Q3

Practice Set 3.4, Q4

Practice Set 3.4, Q5

Practice Set 3.4, Q6

Problem Set 3

Problem Set 3, Q1(1)

Problem Set 3, Q1(2)

Problem Set 3, Q1(3)

Problem Set 3, Q1(4)

Problem Set 3, Q1(5)

Problem Set 3, Q1(6)

Problem Set 3, Q1(7)

Problem Set 3, Q1(8)

Problem Set 3, Q1(9)

Problem Set 3, Q1(10)

Problem Set 3, Q2

Problem Set 3, Q3

Problem Set 3, Q4

Problem Set 3, Q5

Problem Set 3, Q6

Problem Set 3, Q7

Problem Set 3, Q8

Problem Set 3, Q9

Problem Set 3, Q10

Problem Set 3, Q14