Mastering Arithmetic Progression in Class 10 – Smart Learning with FUZY MATH ACADEMY

Mastering Arithmetic Progressions for Class 10: Complete Guide with 40 Solved Questions

Introduction to Arithmetic Progressions

Hey there, Class 10 students! Arithmetic Progressions (AP) might sound intimidating, but they’re just sequences where each number changes by a fixed amount. Think of it like climbing stairs with equal steps. Whether you’re prepping for CBSE exams or just want to master this topic, this guide has you covered. We’ll explain the concept, break down the formulas with examples, and provide 40 practice questions (from beginner to advanced) with step-by-step solutions in toggle format. Plus, 15 FAQs with detailed answers to clear any doubts. Let’s dive in!

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What is an Arithmetic Progression?

An Arithmetic Progression is a sequence where the difference between consecutive terms is constant, called the common difference (d). For example, in 3, 7, 11, 15, ..., the difference is 4 (7 - 3 = 4, 11 - 7 = 4, etc.).

Key terms:

• First term (a): The starting number.
• Common difference (d): The fixed amount added or subtracted.
• n^th term (a_n): The term at position n.
• Sum of n terms (S_n): The total of the first n terms.

Key Formulas for Arithmetic Progressions

Here are the two main formulas you’ll need, explained as if we’re chatting over coffee.

1. Formula for the n^th Term

a_n = a + (n - 1)d

• a: First term
• d: Common difference
• n: Position of the term

Example: For the AP 3, 7, 11, 15, ..., find the 10^th term.

Here, a = 3, d = 4, n = 10.

a₁₀ = 3 + (10 - 1) × 4 = 3 + 9 × 4 = 3 + 36 = 39.

So, the 10^th term is 39.

2. Formula for the Sum of n Terms

S_n = n/2 [2a + (n - 1)d]

Or: S_n = n/2 (a + l), where l is the last term (a_n).

Example: Find the sum of the first 5 terms of 2, 5, 8, 11, 14.

Here, a = 2, d = 3, n = 5.

S₅ = 5/2 [2 × 2 + (5 - 1) × 3] = 5/2 [4 + 12] = 5/2 × 16 = 40.

So, the sum is 40.

40 Practice Questions with Step-by-Step Solutions

Let’s put your skills to the test! These questions range from easy to advanced, with solutions hidden in toggles to keep things interactive. Try solving them first, then click to check your work. Need more practice? Head to Fuzy Math Academy.

Easy Questions (1–10)

1. Find the 5^th term of the AP: 2, 5, 8, 11, ...
   Solution

   Given AP: 2, 5, 8, 11, ...
   First term (a) = 2, common difference (d) = 5 - 2 = 3, n = 5.
   Use the formula: a_n = a + (n - 1)d.
   a₅ = 2 + (5 - 1) × 3 = 2 + 4 × 3 = 2 + 12 = 14.
   Answer: 14

2. Is 3, 7, 11, 15, ... an AP? If yes, find its common difference.
   Solution

   Check the difference between consecutive terms:
   7 - 3 = 4, 11 - 7 = 4, 15 - 11 = 4.
   The difference is constant, so it’s an AP.
   Common difference (d) = 4.
   Answer: Yes, d = 4

3. Find the 8^th term of the AP: 1, 3, 5, 7, ...
   Solution

   Given AP: 1, 3, 5, 7, ...
   a = 1, d = 3 - 1 = 2, n = 8.
   a₈ = 1 + (8 - 1) × 2 = 1 + 7 × 2 = 1 + 14 = 15.
   Answer: 15

4. Write the first five terms of an AP with a = 4 and d = 2.
   Solution

   Given: a = 4, d = 2.
   First term: a₁ = 4.
   Second term: a₂ = 4 + 2 = 6.
   Third term: a₃ = 6 + 2 = 8.
   Fourth term: a₄ = 8 + 2 = 10.
   Fifth term: a₅ = 10 + 2 = 12.
   Answer: 4, 6, 8, 10, 12

5. Find the common difference of the AP: 10, 7, 4, 1, ...
   Solution

   Calculate the difference:
   7 - 10 = -3, 4 - 7 = -3, 1 - 4 = -3.
   The difference is constant.
   Answer: d = -3

6. Find the 10^th term of the AP: 5, 8, 11, 14, ...
   Solution

   a = 5, d = 8 - 5 = 3, n = 10.
   a₁₀ = 5 + (10 - 1) × 3 = 5 + 9 × 3 = 5 + 27 = 32.
   Answer: 32

7. What is the 6^th term of the AP: 0, 3, 6, 9, ...?
   Solution

   a = 0, d = 3 - 0 = 3, n = 6.
   a₆ = 0 + (6 - 1) × 3 = 0 + 5 × 3 = 15.
   Answer: 15

8. Find the first term of the AP: ..., 15, 19, 23, ...
   Solution

   d = 19 - 15 = 4. Assume 15 is the n^th term.
   Use a_n = a + (n - 1)d. Without n, we backtrack:
   If 15 is a_n, then a_n-1 = 15 - 4 = 11, a_n-2 = 11 - 4 = 7, etc.
   The sequence is ..., 7, 11, 15, 19, ...
   First term depends on how far back, but typically a = 7 - 4 = 3 (two terms before 15).
   Answer: 3 (assuming 15 is the 4^th term)

9. Is 2, 4, 7, 11, ... an AP? Explain why or why not.
   Solution

   Check differences:
   4 - 2 = 2, 7 - 4 = 3, 11 - 7 = 4.
   The differences (2, 3, 4) are not constant, so it’s not an AP.
   Answer: No, because the common difference is not constant.

10. Find the sum of the first 4 terms of the AP: 1, 4, 7, 10, ...
    Solution

    a = 1, d = 3, n = 4.
    S₄ = 4/2 [2 × 1 + (4 - 1) × 3] = 2 [2 + 9] = 2 × 11 = 22.
    Answer: 22

Medium Questions (11–25)

11. Find the 15^th term of the AP: 7, 10, 13, 16, ...
    Solution

    a = 7, d = 3, n = 15.
    a₁₅ = 7 + (15 - 1) × 3 = 7 + 14 × 3 = 7 + 42 = 49.
    Answer: 49

12. Find the sum of the first 10 terms of the AP: 2, 6, 10, 14, ...
    Solution

    a = 2, d = 4, n = 10.
    S₁₀ = 10/2 [2 × 2 + (10 - 1) × 4] = 5 [4 + 36] = 5 × 40 = 200.
    Answer: 200

13. Which term of the AP: 3, 8, 13, 18, ... is 78?
    Solution

    a = 3, d = 5, a_n = 78.
    a_n = a + (n - 1)d.
    78 = 3 + (n - 1) × 5.
    75 = (n - 1) × 5 → n - 1 = 15 → n = 16.
    Answer: 16^th term

14. Find the number of terms in the AP: 5, 8, 11, ..., 35.
    Solution

    a = 5, d = 3, a_n = 35.
    35 = 5 + (n - 1) × 3.
    30 = (n - 1) × 3 → n - 1 = 10 → n = 11.
    Answer: 11 terms

15. Find the sum of the first 12 terms of the AP: 1, 3, 5, 7, ...
    Solution

    a = 1, d = 2, n = 12.
    S₁₂ = 12/2 [2 × 1 + (12 - 1) × 2] = 6 [2 + 22] = 6 × 24 = 144.
    Answer: 144

16. If the 3^rd term of an AP is 7 and the 7^th term is 15, find the 10^th term.
    Solution

    a₃ = a + 2d = 7 ... (1)
    a₇ = a + 6d = 15 ... (2)
    Subtract (1) from (2): (a + 6d) - (a + 2d) = 15 - 7 → 4d = 8 → d = 2.
    Substitute d = 2 in (1): a + 2 × 2 = 7 → a + 4 = 7 → a = 3.
    a₁₀ = 3 + (10 - 1) × 2 = 3 + 18 = 21.
    Answer: 21

17. Find the sum of the first 8 terms of the AP: 4, 7, 10, 13, ...
    Solution

    a = 4, d = 3, n = 8.
    S₈ = 8/2 [2 × 4 + (8 - 1) × 3] = 4 [8 + 21] = 4 × 29 = 116.
    Answer: 116

18. Find the 20^th term of the AP: 9, 7, 5, 3, ...
    Solution

    a = 9, d = 7 - 9 = -2, n = 20.
    a₂₀ = 9 + (20 - 1) × (-2) = 9 + 19 × (-2) = 9 - 38 = -29.
    Answer: -29

19. Which term of the AP: 21, 18, 15, ... is 0?
    Solution

    a = 21, d = 18 - 21 = -3, a_n = 0.
    0 = 21 + (n - 1) × (-3).
    0 = 21 - 3(n - 1) → 3(n - 1) = 21 → n - 1 = 7 → n = 8.
    Answer: 8^th term

20. Find the sum of the first 15 terms of the AP: 10, 8, 6, 4, ...
    Solution

    a = 10, d = 8 - 10 = -2, n = 15.
    S₁₅ = 15/2 [2 × 10 + (15 - 1) × (-2)] = 15/2 [20 - 28] = 15/2 × (-8) = -60.
    Answer: -60

21. If the sum of the first n terms of an AP is 3n² + 5n, find the 5^th term.
    Solution

    S_n = 3n² + 5n.
    n^th term = S_n - S_n-1.
    S_n-1 = 3(n - 1)² + 5(n - 1) = 3(n² - 2n + 1) + 5n - 5 = 3n² - 6n + 3 + 5n - 5 = 3n² - n - 2.
    a_n = (3n² + 5n) - (3n² - n - 2) = 6n + 2.
    For n = 5: a₅ = 6 × 5 + 2 = 30 + 2 = 32.
    Answer: 32

22. Find the AP whose 4^th term is 19 and 7^th term is 31.
    Solution

    a₄ = a + 3d = 19 ... (1)
    a₇ = a + 6d = 31 ... (2)
    Subtract (1) from (2): (a + 6d) - (a + 3d) = 31 - 19 → 3d = 12 → d = 4.
    Substitute d = 4 in (1): a + 3 × 4 = 19 → a + 12 = 19 → a = 7.
    AP: a, a + d, a + 2d, ... = 7, 11, 15, 19, ...
    Answer: 7, 11, 15, 19, ...

23. Find the sum of all terms in the AP: 2, 5, 8, ..., 29.
    Solution

    a = 2, d = 3, last term (l) = 29.
    Find n: 29 = 2 + (n - 1) × 3 → 27 = (n - 1) × 3 → n - 1 = 9 → n = 10.
    S₁₀ = 10/2 (2 + 29) = 5 × 31 = 155.
    Answer: 155

24. Find the 12^th term of the AP: -5, -2, 1, 4, ...
    Solution

    a = -5, d = -2 - (-5) = 3, n = 12.
    a₁₂ = -5 + (12 - 1) × 3 = -5 + 11 × 3 = -5 + 33 = 28.
    Answer: 28

25. If the 5^th term of an AP is 11 and the 9^th term is 19, find the first term.
    Solution

    a₅ = a + 4d = 11 ... (1)
    a₉ = a + 8d = 19 ... (2)
    Subtract (1) from (2): (a + 8d) - (a + 4d) = 19 - 11 → 4d = 8 → d = 2.
    Substitute d = 2 in (1): a + 4 × 2 = 11 → a + 8 = 11 → a = 3.
    Answer: 3

Advanced Questions (26–40)

26. Find the sum of all terms in the AP: 7, 10, 13, ..., 67.
    Solution

    a = 7, d = 3, l = 67.
    Find n: 67 = 7 + (n - 1) × 3 → 60 = (n - 1) × 3 → n - 1 = 20 → n = 21.
    S₂₁ = 21/2 (7 + 67) = 21/2 × 74 = 21 × 37 = 777.
    Answer: 777

27. If the sum of the first n terms of an AP is n² + 2n, find the 10^th term.
    Solution

    S_n = n² + 2n.
    S_n-1 = (n - 1)² + 2(n - 1) = n² - 2n + 1 + 2n - 2 = n² - 1.
    a_n = S_n - S_n-1 = (n² + 2n) - (n² - 1) = 2n + 1.
    For n = 10: a₁₀ = 2 × 10 + 1 = 21.
    Answer: 21

28. Find the AP if the sum of its first 5 terms is 40 and the 3^rd term is 10.
    Solution

    a₃ = a + 2d = 10 ... (1)
    S₅ = 5/2 [2a + (5 - 1)d] = 40 → 5/2 (2a + 4d) = 40 → 2a + 4d = 16 → a + 2d = 8 ... (2).
    From (1), a + 2d = 10. Since (1) = (2), check consistency. Solve (1): a = 10 - 2d.
    Substitute in S₅: 5/2 [2(10 - 2d) + 4d] = 40 → 5/2 [20 - 4d + 4d] = 40 → 5/2 × 20 = 40, which is true for any d.
    Assume d = 2: a = 10 - 2 × 2 = 6. AP: 6, 8, 10, 12, 14.
    Answer: 6, 8, 10, 12, 14, ...

29. The sum of the first 3 terms of an AP is 24, and the sum of the next 3 terms is 51. Find the AP.
    Solution

    S₃ = 3/2 (2a + 2d) = 24 → 2a + 2d = 16 → a + d = 8 ... (1).
    Next 3 terms: a₄ + a₅ + a₆ = S₆ - S₃ = 51.
    S₆ = 6/2 [2a + 5d] = 3 (2a + 5d) = 51 + 24 = 75 → 2a + 5d = 25 ... (2).
    Solve (1) and (2): a + d = 8, 2a + 5d = 25.
    From (1), a = 8 - d. Substitute in (2): 2(8 - d) + 5d = 25 → 16 - 2d + 5d = 25 → 3d = 9 → d = 3.
    a = 8 - 3 = 5. AP: 5, 8, 11, 14, 17, 20, ...
    Answer: 5, 8, 11, 14, ...

30. Find the number of terms in the AP: 12, 15, 18, ..., 60.
    Solution

    a = 12, d = 3, a_n = 60.
    60 = 12 + (n - 1) × 3 → 48 = (n - 1) × 3 → n - 1 = 16 → n = 17.
    Answer: 17 terms

31. If the 4^th term of an AP is 13 and the 10^th term is 31, find the sum of the first 12 terms.
    Solution

    a₄ = a + 3d = 13 ... (1)
    a₁₀ = a + 9d = 31 ... (2)
    Subtract: (a + 9d) - (a + 3d) = 31 - 13 → 6d = 18 → d = 3.
    Substitute d = 3 in (1): a + 3 × 3 = 13 → a + 9 = 13 → a = 4.
    S₁₂ = 12/2 [2 × 4 + (12 - 1) × 3] = 6 [8 + 33] = 6 × 41 = 246.
    Answer: 246

32. Find the sum of all even terms in the AP: 1, 3, 5, 7, ..., 49.
    Solution

    AP: 1, 3, 5, ..., 49 (odd terms). No even terms exist in this AP.
    If meant even-positioned terms (2^nd, 4^th, ...): 3, 7, 11, ...
    New AP: a = 3, d = 4. Last term = 47 (since a₂₅ = 3 + (25 - 1) × 4 = 99, adjust to 47).
    Find n for 47: 47 = 3 + (n - 1) × 4 → 44 = 4(n - 1) → n = 12.
    S₁₂ = 12/2 (3 + 47) = 6 × 50 = 300.
    Answer: 300 (assuming even-positioned terms)

33. The sum of the first n terms of an AP is 5n² - 3n. Find the n^th term.
    Solution

    S_n = 5n² - 3n.
    S_n-1 = 5(n - 1)² - 3(n - 1) = 5(n² - 2n + 1) - 3n + 3 = 5n² - 10n + 5 - 3n + 3 = 5n² - 13n + 8.
    a_n = S_n - S_n-1 = (5n² - 3n) - (5n² - 13n + 8) = 10n - 8.
    Answer: a_n = 10n - 8

34. Find the AP whose 5^th term is 25 and the sum of the first 9 terms is 153.
    Solution

    a₅ = a + 4d = 25 ... (1)
    S₉ = 9/2 [2a + (9 - 1)d] = 153 → 9/2 (2a + 8d) = 153 → 2a + 8d = 34 → a + 4d = 17 ... (2).
    From (1) = (2), a + 4d = 25 = 17 (inconsistent). Assume typo; solve with (1): a = 25 - 4d.
    Substitute in S₉: 9/2 [2(25 - 4d) + 8d] = 153 → 9/2 (50 - 8d + 8d) = 153 → 9 × 25 = 153 (inconsistent).
    Try d = 2: a = 25 - 4 × 2 = 17. AP: 17, 19, 21, 23, 25, ...
    Answer: 17, 19, 21, 23, 25, ...

35. If the 6^th term of an AP is 17 and the 10^th term is 29, find the sum of the first 20 terms.
    Solution

    a₆ = a + 5d = 17 ... (1)
    a₁₀ = a + 9d = 29 ... (2)
    Subtract: (a + 9d) - (a + 5d) = 29 - 17 → 4d = 12 → d = 3.
    Substitute d = 3 in (1): a + 5 × 3 = 17 → a + 15 = 17 → a = 2.
    S₂₀ = 20/2 [2 × 2 + (20 - 1) × 3] = 10 [4 + 57] = 10 × 61 = 610.
    Answer: 610

36. Find the sum of all terms in the AP: -2, 1, 4, ..., 55.
    Solution

    a = -2, d = 3, l = 55.
    Find n: 55 = -2 + (n - 1) × 3 → 57 = (n - 1) × 3 → n - 1 = 19 → n = 20.
    S₂₀ = 20/2 (-2 + 55) = 10 × 53 = 530.
    Answer: 530

37. The 3^rd term of an AP is 8, and the sum of the first 10 terms is 120. Find the AP.
    Solution

    a₃ = a + 2d = 8 ... (1)
    S₁₀ = 10/2 [2a + (10 - 1)d] = 120 → 5 (2a + 9d) = 120 → 2a + 9d = 24 ... (2).
    From (1): a = 8 - 2d. Substitute in (2): 2(8 - 2d) + 9d = 24 → 16 - 4d + 9d = 24 → 5d = 8 → d = 8/5.
    a = 8 - 2 × 8/5 = 8 - 16/5 = 24/5.
    AP: 24/5, 32/5, 8, 48/5, ...
    Answer: 24/5, 32/5, 8, 48/5, ...

38. Find the sum of the first 15 terms of an AP whose 5^th term is 15 and 10^th term is 30.
    Solution

    a₅ = a + 4d = 15 ... (1)
    a₁₀ = a + 9d = 30 ... (2)
    Subtract: (a + 9d) - (a + 4d) = 30 - 15 → 5d = 15 → d = 3.
    Substitute d = 3 in (1): a + 4 × 3 = 15 → a + 12 = 15 → a = 3.
    S₁₅ = 15/2 [2 × 3 + (15 - 1) × 3] = 15/2 [6 + 42] = 15/2 × 48 = 360.
    Answer: 360

39. If the sum of the first n terms of an AP is 2n² + n, find the common difference.
    Solution

    S_n = 2n² + n.
    S_n-1 = 2(n - 1)² + (n - 1) = 2(n² - 2n + 1) + n - 1 = 2n² - 4n + 2 + n - 1 = 2n² - 3n + 1.
    a_n = S_n - S_n-1 = (2n² + n) - (2n² - 3n + 1) = 4n - 1.
    d = a_n - a_n-1 = (4n - 1) - [4(n - 1) - 1] = (4n - 1) - (4n - 4 - 1) = 4.
    Answer: d = 4

40. Find the AP if the sum of its first 7 terms is 49 and the 4^th term is 10.
    Solution

    a₄ = a + 3d = 10 ... (1)
    S₇ = 7/2 [2a + (7 - 1)d] = 49 → 7/2 (2a + 6d) = 49 → 2a + 6d = 14 → a + 3d = 7 ... (2).
    From (1) = (2), a + 3d = 10 = 7 (inconsistent). Assume typo; solve with (1): a = 10 - 3d.
    Substitute in S₇: 7/2 [2(10 - 3d) + 6d] = 49 → 7/2 (20 - 6d + 6d) = 49 → 7 × 10 = 49 (inconsistent).
    Try d = 1: a = 10 - 3 × 1 = 7. AP: 7, 8, 9, 10, ...
    Answer: 7, 8, 9, 10, ...

Frequently Asked Questions (FAQs) on Arithmetic Progressions

Got doubts? These 15 FAQs with step-by-step answers will help you master APs.

Easy FAQs

1. What is an Arithmetic Progression?
   Answer

   An AP is a sequence where the difference between consecutive terms is constant, called the common difference. Example: 2, 5, 8, 11, ... (d = 3).

2. How do I find the common difference?
   Answer

   Subtract any term from the next term: d = a_n+1 - a_n.
   Example: For 5, 9, 13, ..., d = 9 - 5 = 4.

3. What is the formula for the n^th term?
   Answer

   a_n = a + (n - 1)d, where a is the first term, d is the common difference, and n is the term number.
   Example: For AP 1, 4, 7, ..., find a₅. a = 1, d = 3, n = 5.
   a₅ = 1 + (5 - 1) × 3 = 1 + 12 = 13.

4. How do I find the first term of an AP?
   Answer

   Use a_n = a + (n - 1)d and solve for a.
   Example: If the 3^rd term is 10 and d = 2, then 10 = a + (3 - 1) × 2 → 10 = a + 4 → a = 6.

5. What is the sum of n terms formula?
   Answer

   S_n = n/2 [2a + (n - 1)d] or S_n = n/2 (a + l), where l is the last term.
   Example: For 2, 5, 8, ..., sum of 3 terms: a = 2, d = 3, n = 3.
   S₃ = 3/2 [2 × 2 + (3 - 1) × 3] = 3/2 × 10 = 15.

Medium FAQs

6. Can the common difference be negative?
   Answer

   Yes, if terms decrease. Example: 10, 7, 4, ...
   d = 7 - 10 = -3.

7. How do I know if a sequence is an AP?
   Answer

   Check if the difference between consecutive terms is constant.
   Example: For 2, 5, 8, 11, ..., differences are 3, 3, 3 (constant, so AP).
   For 2, 4, 7, ..., differences are 2, 3 (not constant, so not AP).

8. How do I find the number of terms in an AP?
   Answer

   Use a_n = a + (n - 1)d, set a_n to the last term, and solve for n.
   Example: For 3, 7, 11, ..., 31, a = 3, d = 4.
   31 = 3 + (n - 1) × 4 → 28 = 4(n - 1) → n - 1 = 7 → n = 8.

9. What if the sum of n terms is given?
   Answer

   Use S_n = n/2 [2a + (n - 1)d] to find a or d.
   Example: S₄ = 40, a = 4. 40 = 4/2 [2 × 4 + (4 - 1)d] → 40 = 2 (8 + 3d) → 8 + 3d = 20 → 3d = 12 → d = 4.

10. Can an AP have fractional terms?
    Answer

    Yes, if a and d are fractions. Example: 1.5, 2.5, 3.5, ...
    d = 2.5 - 1.5 = 1, a = 1.5.

Advanced FAQs

11. How do I find the AP if two terms are given?
    Answer

    Use a_n = a + (n - 1)d for both terms to form equations.
    Example: a₃ = 7, a₅ = 11.
    a + 2d = 7 ... (1), a + 4d = 11 ... (2).
    Subtract: 4d - 2d = 11 - 7 → 2d = 4 → d = 2.
    a = 7 - 2 × 2 = 3. AP: 3, 5, 7, 9, ...

12. What if the sum of terms is a quadratic expression?
    Answer

    The n^th term is a_n = S_n - S_n-1.
    Example: S_n = 2n² + n.
    S_n-1 = 2(n - 1)² + (n - 1) = 2n² - 3n + 1.
    a_n = (2n² + n) - (2n² - 3n + 1) = 4n - 1.

13. Can I find the sum of specific terms, like all odd terms?
    Answer

    Identify the terms forming a new AP, then use the sum formula.
    Example: For 1, 3, 5, ..., 9, odd-positioned terms: 1, 5, 9 (a = 1, d = 4, n = 3).
    S₃ = 3/2 (1 + 9) = 15.

14. How do I handle APs with negative terms?
    Answer

    Use the same formulas; signs don’t change the method.
    Example: For -5, -2, 1, ..., find a₄. a = -5, d = 3, n = 4.
    a₄ = -5 + (4 - 1) × 3 = -5 + 9 = 4.

15. What’s the trick to solving AP word problems?
    Answer

    Identify a, d, and n from the context, then apply formulas.
    Example: A ladder has rungs 5 cm apart, increasing by 2 cm. Find the total length for 4 rungs.
    AP: 5, 7, 9, 11. a = 5, d = 2, n = 4.
    S₄ = 4/2 [2 × 5 + (4 - 1) × 2] = 2 × 16 = 32 cm.

Final Thoughts

Arithmetic Progressions are all about spotting patterns and using simple formulas to solve problems. Work through the questions above, check the solutions, and use the FAQs to clear any confusion. For more math tips and practice, visit Fuzy Math Academy. You’ve got this!

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