Arithmetic Progression Questions: The Complete Guide

Solution:
a = 45, d = 39-45 = -6, Sₙ = 180
Sₙ = n/2 [2a + (n-1)d]
180 = n/2 [90 + (n-1)(-6)]
360 = n[90 - 6n + 6]
360 = n[96 - 6n]
360 = 96n - 6n²
6n² - 96n + 360 = 0
n² - 16n + 60 = 0
(n-10)(n-6) = 0
n = 6 or n = 10

Example 19: First Term from Given Term and Difference

Question:
In an AP, if d = -4, and the seventh term is 4, find the first term.

Solution:
a₇ = a + 6d = 4
d = -4
a + 6×(-4) = 4
a - 24 = 4
a = 28

Example 20: Four Consecutive AP Numbers with a Condition

Question:
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and last term to the product of the two middle terms is 7:15. Find the numbers.

Solution:
Let numbers be a-3d, a-d, a+d, a+3d
Sum: a-3d + a-d + a+d + a+3d = 4a = 32 ⇒ a = 8
Product of ends: (a-3d)(a+3d) = a²-9d²
Product of middles: (a-d)(a+d) = a²-d²
(a²-9d²)/(a²-d²) = 7/15
Cross-multiplied: 15(a²-9d²) = 7(a²-d²)
15a²-135d² = 7a²-7d²
8a² = 128d² ⇒ a² = 16d² ⇒ a = 8, d = ±2
So, numbers: 8-6=2, 8-2=6, 8+2=10, 8+6=14

Example 21: Common Difference from AP Terms

Question:
What is the common difference of an AP in which a₂₁ - a₇ = 84?

Solution:
a₂₁ = a + 20d
a₇ = a + 6d
a₂₁ - a₇ = (a + 20d) - (a + 6d) = 14d = 84
d = 6

Example 22: First Negative Term in a Fractional AP

Question:
Which term of the progression 20, 19¼, 18½, 17¾, … is the first negative term?

Solution:
a = 20, d = 19.25 - 20 = -0.75
aₙ < 0 ⇒ 20 + (n-1)(-0.75) < 0 20 - 0.75n + 0.75 < 0 20.75 < 0.75n n > 27.67
So, the 28th term is the first negative term.

Example 23: nth Term from Ratio of Sums

Question:
If the ratio of the sum of the first n terms of two APs is (7n + 1):(4n + 27), what is the ratio of their 9th terms?

Solution:
Let Sₙ₁/Sₙ₂ = (7n+1)/(4n+27)
Sₙ = n/2 [2a + (n-1)d]
a₉ = a + 8d
Difference in sum for 9 and 8 terms gives the 9th term.
Ratio for n=9:
S₉₁/S₉₂ = (7×9+1)/(4×9+27) = (63+1)/(36+27) = 64/63
So, the ratio of the 9th terms is also 64/63.

Example 24: AP Term from Two Given Terms

Question:
If the 8th term of an AP is zero, and the 12th term is 20, find the first term and the common difference.

Solution:
a₈ = a + 7d = 0
a₁₂ = a + 11d = 20
Subtract: (a + 11d) - (a + 7d) = 20 - 0 ⇒ 4d = 20 ⇒ d = 5
a = 0 - 7×5 = -35

Example 25: Finding the 19th Term

Question:
Find the 19th term of the AP: 5, 11, 17, 23, …

Solution:
a = 5, d = 6
a₁₉ = 5 + (19-1)×6 = 5 + 108 = 113

Example 26: Sum of First 30 Even Natural Numbers

Question:
What is the sum of the first 30 even natural numbers?

Solution:
First even number: 2
d = 2
a₃₀ = 2 + 29×2 = 60
Sum = n/2 (a₁ + aₙ) = 30/2 (2+60) = 15×62 = 930

Example 27: Which Term is 80 More than the 20th Term (Alternate AP)?

Question:
Which term of the AP 13, 18, 23, … is 80 more than its 20th term?
(Repeat for reinforcement)

Solution:
Already solved in Example 9. The answer is the 36th term.

Example 28: AP with Variable First Term and Difference

Question:
The first term of an AP is p and its common difference is q. Find its 10th term.

Solution:
a = p, d = q
a₁₀ = p + (10-1)q = p + 9q

Example 29: Number of Terms for Zero Sum

Question:
How many terms of the AP 18, 16, 14, … must be taken so that their sum is zero?

Solution:
a = 18, d = -2, Sₙ = 0
Sₙ = n/2 [2a + (n-1)d]
0 = n/2 [36 + (n-1)(-2)]
0 = n/2 [36 - 2n + 2]
0 = n/2 [38 - 2n]
n = 0 or 38 - 2n = 0 ⇒ n = 19

Example 30: Proving a Relationship Between Terms

Question:
The 4th term of an AP is zero. Prove that the 25th term is three times its 11th term.

Solution:
Let a₄ = 0 ⇒ a + 3d = 0 ⇒ a = -3d
a₂₅ = a + 24d = -3d + 24d = 21d
a₁₁ = a + 10d = -3d + 10d = 7d
a₂₅ = 3 × a₁₁ (since 21d = 3 × 7d)
Hence proved.

Test your knowledge and problem-solving skills with the following arithmetic progression practice questions. These are ideal for preparing for arithmetic progression quiz sections and can also be found in arithmetic progression questions and answers pdf resources. Work through each one carefully and check your answers after attempting them all.

Basic Level

1. Find the common difference of the AP: 4, 9, 14, 19, …
2. What is the 12th term of the AP: 6, 10, 14, 18, …?
3. Write the first five terms of the AP whose first term is 7 and common difference is -3.
4. Which term of the AP: 21, 18, 15, … is -81?
5. Find the sum of the first 10 terms of the AP: 5, 8, 11, …
6. If the 5th term of an AP is 20 and the common difference is 4, what is the first term?
7. The nth term of an AP is 2n + 3. Find the 8th term.
8. Find the 25th term of the AP: 100, 95, 90, …
9. If the sum of the first n terms of an AP is 3n² + 2n, find its 5th term.
10. Write the next three terms: 2, 6, 10, 14, …

Intermediate Level

11. Which term of the AP: 17, 14, 11, … will be -40?
12. Find the sum of the first 20 terms of the AP: 1, 3, 5, …
13. If the 7th term of an AP is 13 and the 12th term is 28, find the common difference.
14. In the AP: 50, 47, 44, …, how many terms are needed for the sum to be 150?
15. Find the number of terms in the AP: 7, 13, 19, …, 85.
16. Find the sum of all two-digit numbers which are divisible by 5.
17. If the sum of the first n terms of an AP is 5n, what is its nth term?
18. Which term of the AP: 13, 8, 3, … is -27?
19. If the 3rd term of an AP is 12 and the 8th term is 27, what is the common difference?
20. Find the sum of the first 15 terms of the AP: 10, 7, 4, …

Application and Word Problems

21. A student saves Rs 10 in the first week and increases his savings by Rs 5 every week. How much will he save in the 12th week?
22. The sum of the first 8 terms of an AP is 100. If the first term is 6, find the common difference.
23. The sum of three consecutive terms of an AP is 24, and their product is 440. Find the terms.
24. If the 2nd term of an AP is 7 and the 6th term is 23, find the first term and the common difference.
25. The sum of the first n terms of an AP is 2n² + n. Find its 9th term.
26. In an AP, the 1st term is 2 and the 5th term is 14. Find the sum of the first 10 terms.
27. The 10th term of an AP is 50 and the common difference is 3. What is the first term?
28. If the sum of the first 20 terms of an AP is 400 and the first term is 5, find the common difference.
29. Which term of the AP: 29, 25, 21, … is -15?
30. The 4th term of an AP is 0. Prove that the 25th term is three times its 11th term.

Advanced and Exam-Oriented

31. If the sum of the first n terms of an AP is n² + 2n, find its 7th term.
32. In an AP, the 8th term is 37 and the 15th term is 65. Find the common difference and the first term.
33. The sum of the first 16 terms of an AP is 352. If the first term is 7, find the common difference.
34. The sum of the first 30 terms of an AP is 1500. If the last term is 100, find the first term.
35. The 2nd and 5th terms of an AP are 10 and 19, respectively. Find the 20th term.
36. Find the sum of the first 25 terms of the AP: 7, 10, 13, …
37. If the 12th term of an AP is 50 and the sum of the first 12 terms is 312, what is the first term?
38. The sum of the first n terms of an AP is 6n - n². Find the AP.
39. In an AP, the sum of the first n terms is 5n² + 3n. Find its common difference.
40. Which term of the AP 4, 7, 10, … is 100?

Mixed Practice

41. Find the sum of the first 50 natural numbers.
42. The sum of the first n terms of an AP is 2n² + 3n. Find the 10th term.
43. Which term of the AP: 20, 18.5, 17, … is the first negative term?
44. If the 6th term of an AP is 22 and the 15th term is 49, find the common difference.
45. The sum of the first 5 terms of an AP is 40 and the sum of the next 5 terms is 90. Find the AP.
46. If a, b, c are in AP and a + b + c = 18, find the value of b.
47. The sum of three consecutive terms of an AP is 27, and their product is 504. Find the terms.
48. If the sum of the first 7 terms of an AP is 119, and the first term is 6, find the common difference.
49. Find the 20th term of the AP: 15, 12, 9, …
50. Which term of the AP 7, 13, 19, … is 85?

Challenging and Conceptual

51. The sum of the first n terms of an AP is 3n² + n. Find the 10th term.
52. In an AP, the sum of the first 8 terms is 100, and the sum of the next 8 terms is 340. Find the first term and the common difference.
53. Which term of the AP: 100, 97, 94, … is the first negative term?
54. If the 5th term of an AP is 0, and the 9th term is 8, find the AP.
55. The sum of the first 30 terms of an AP is 750, and the last term is 49. Find the first term.
56. Find the sum of all odd numbers between 1 and 100.
57. If the sum of the first n terms of an AP is 4n² + n, what is the 11th term?
58. The 7th term of an AP is 20 and the 13th term is 38. Find the AP.
59. Which term of the AP: 12, 9, 6, … is -33?
60. If the sum of the first 16 terms of an AP is 400, and the first term is 4, find the common difference.

Real-World and Word Problems

61. A person borrows Rs 5000 and agrees to pay back in 10 monthly installments, each installment being Rs 50 more than the previous one. What is the value of the first and last installment?
62. The number of seats in each row of a theater increases by 4 from row to row. If the first row has 20 seats, how many seats are there in the 15th row?
63. A sum of Rs 1200 is to be distributed among 12 students so that each gets Rs 10 more than the previous student. Find the share of the first and last student.
64. The cost of a taxi ride is Rs 50 for the first kilometer and increases by Rs 10 for each subsequent kilometer. What is the cost for a 12 km ride?
65. A school awards scholarships increasing by Rs 500 each year. If the first award is Rs 2000, what is the amount of the 10th award?
66. A ladder has 20 rungs, each 30 cm apart. What is the distance from the first rung to the last?
67. The population of a town increases by 2% each year. If the current population is 5000, what will it be after 10 years (use AP approximation)?
68. A pile of bricks has 25 layers. The top layer has 10 bricks, and each lower layer has 2 more bricks than the layer above. How many bricks are there in total?
69. In an auditorium, there are 30 rows of seats. The first row has 8 seats, and each subsequent row has 2 more seats. How many seats are there in total?
70. A person starts jogging with 1 km on the first day and increases the distance by 0.5 km each day. How far will he jog on the 20th day?.

Solutions

Basic Level

1. Solution:
   d = 9 - 4 = 5
2. Solution:
   a = 6, d = 4
   a₁₂ = a + (12-1)d = 6 + 11×4 = 6 + 44 = 50
3. Solution:
   7, 4, 1, -2, -5
4. Solution:
   a = 21, d = -3
   aₙ = 21 + (n-1)(-3) = -81
   -81 = 21 - 3n + 3
   -81 = 24 - 3n
   -105 = -3n
   n = 35
5. Solution:
   a = 5, d = 3, n = 10
   S₁₀ = n/2 [2a + (n-1)d] = 10/2 [10 + 9×3] = 5[10 + 27] = 5×37 = 185
6. Solution:
   a₅ = a + 4d = 20
   a + 16 = 20
   a = 4
7. Solution:
   a₈ = 2×8 + 3 = 16 + 3 = 19
8. Solution:
   a = 100, d = -5
   a₂₅ = 100 + (25-1)(-5) = 100 + 24×(-5) = 100 - 120 = -20
9. Solution:
   a₅ = S₅ - S₄ = [3×25 + 2×5] - [3×16 + 2×4] = (75+10)-(48+8) = 85-56 = 29
10. Solution:
    d = 4
    Next terms: 18, 22, 26

Intermediate Level

11. Solution:
    a = 17, d = -3
    aₙ = 17 + (n-1)(-3) = -40
    -40 = 17 - 3n + 3
    -40 = 20 - 3n
    -60 = -3n
    n = 20
12. Solution:
    a = 1, d = 2, n = 20
    S₂₀ = 20/2 [2×1 + 19×2] = 10[2 + 38] = 10×40 = 400
13. Solution:
    a₇ = a + 6d = 13
    a₁₂ = a + 11d = 28
    Subtract: (a + 11d) - (a + 6d) = 28 - 13 → 5d = 15 → d = 3
14. Solution:
    a = 50, d = -3, Sₙ = 150
    Sₙ = n/2 [2×50 + (n-1)(-3)] = n/2 [100 - 3n + 3] = n/2 [103 - 3n]
    150×2 = n(103 - 3n)
    300 = 103n - 3n²
    3n² - 103n + 300 = 0
    n² - 34.33n + 100 = 0 (approximate)
    Try n = 3: 3² - 34.33×3 + 100 ≈ 9 - 103 + 100 = 6 (not zero)
    Try n = 4: 16 - 137 + 100 = -21 (not zero)
    Try n = 12: 144 - 412 + 100 = -168
    Try n = 25: 625 - 858 + 100 = -133
    Try n = 100: 10000 - 3433 + 100 = 6667
    Alternative: Use quadratic formula:
    3n² - 103n + 300 = 0
    n = [103 ± √(103² - 4×3×300)] / 6
    n = [103 ± √(10609 - 3600)] / 6
    n = [103 ± √7009] / 6
    Since n must be integer, check for calculation error or clarify question.
    (If you want, you can leave it as: n = [103 ± √7009]/6)
15. Solution:
    a = 7, d = 6, aₙ = 85
    85 = 7 + (n-1)×6
    85 - 7 = 6(n-1)
    78 = 6(n-1)
    n-1 = 13 → n = 14
16. Solution:
    Smallest two-digit number divisible by 5: 10
    Largest: 95
    a = 10, d = 5, l = 95
    n = [(95 - 10)/5] + 1 = (85/5) + 1 = 17 + 1 = 18
    S₁₈ = 18/2 (10 + 95) = 9×105 = 945
17. Solution:
    aₙ = Sₙ - Sₙ₋₁ = 5n - 5(n-1) = 5n - 5n + 5 = 5
18. Solution:
    a = 13, d = -5
    aₙ = 13 + (n-1)(-5) = -27
    -27 = 13 - 5n + 5
    -27 = 18 - 5n
    -45 = -5n
    n = 9
19. Solution:
    a₃ = a + 2d = 12
    a₈ = a + 7d = 27
    Subtract: 7d - 2d = 27 - 12 → 5d = 15 → d = 3
20. Solution:
    a = 10, d = -3, n = 15
    S₁₅ = 15/2 [2×10 + 14×(-3)] = 7.5[20 - 42] = 7.5×(-22) = -165

Application and Word Problems

21. Solution:
    a = 10, d = 5, n = 12
    a₁₂ = 10 + (12-1)×5 = 10 + 55 = 65
22. .Solution:
    S₈ = 100, a = 6
    S₈ = 8/2 [2×6 + 7d] = 4[12 + 7d] = 100
    12 + 7d = 25
    7d = 13 → d = 13/7 ≈ 1.857
23. Solution:
    Let terms be a-d, a, a+d
    Sum: (a-d) + a + (a+d) = 3a = 24 → a = 8
    Product: (8-d)×8×(8+d) = 440
    (8-d)(8+d) = 440/8 = 55
    64 - d² = 55 → d² = 9 → d = 3
    Terms: 8-3=5, 8, 8+3=11
24. Solution:
    a₂ = a + d = 7
    a₆ = a + 5d = 23
    Subtract: 5d - d = 23 - 7 → 4d = 16 → d = 4
    a = 7 - 4 = 3
25. Solution:
    a₉ = S₉ - S₈ = [2×81 + 9] - [2×64 + 8] = (162+9)-(128+8)=171-136=35
26. Solution:
    a = 2, a₅ = 2 + 4d = 14 → 4d = 12 → d = 3
    S₁₀ = 10/2 [2×2 + 9×3] = 5[4 + 27]=5×31=155
27. Solution:
    a₁₀ = a + 9d = 50
    a + 27 = 50 → a = 23
28. Solution:
    S₂₀ = 400, a = 5
    S₂₀ = 10 [2×5 + 19d] = 10[10 + 19d] = 400
    10 + 19d = 40 → 19d = 30 → d = 30/19 ≈ 1.579
29. Solution:
    a = 29, d = -4
    aₙ = 29 + (n-1)(-4) = -15
    -15 = 29 - 4n + 4
    -15 = 33 - 4n
    -48 = -4n → n = 12
30. Solution:
    a + 3d = 0 → a = -3d
    a₂₅ = a + 24d = -3d + 24d = 21d
    a₁₁ = a + 10d = -3d + 10d = 7d
    a₂₅ = 3 × a₁₁

Advanced and Exam-Oriented

31. Solution:
    a₇ = S₇ - S₆ = (49+14)-(36+12)=63-48=15
32. Solution:
    a₈ = a+7d = 37
    a₁₅ = a+14d = 65
    Subtract: 7d → 65-37 = 28 → 7d = 28 → d = 4
    a = 37 - 7×4 = 9
33. Solution:
    S₁₆ = 352, a = 7
    S₁₆ = 8[2×7 + 15d] = 8[14 + 15d] = 352
    14 + 15d = 44 → 15d = 30 → d = 2
34. Solution:
    S₃₀ = 1500, a₃₀ = 100
    Sₙ = n/2 (a₁ + aₙ)
    1500 = 15(a₁ + 100)
    a₁ + 100 = 100 → a₁ = 0
35. Solution:
    a₂ = a + d = 10
    a₅ = a + 4d = 19
    4d - d = 19 - 10 → 3d = 9 → d = 3
    a = 10 - 3 = 7
    a₂₀ = 7 + 19×3 = 7 + 57 = 64
36. Solution:
    a = 7, d = 3, n = 25
    S₂₅ = 25/2 [2×7 + 24×3] = 12.5[14 + 72]=12.5×86=1075
37. Solution:
    a₁₂ = a + 11d = 50
    S₁₂ = 6(a + 50) = 312
    a + 50 = 52 → a = 2
38. Solution:
    a₁ = S₁ = 6-1=5
    a₂ = S₂-S₁ = (12-4)-(6-1)=8-5=3
    d = a₂ - a₁ = 3-5 = -2
    AP: 5, 3, 1, -1, …
39. Solution:
    a₁ = S₁ = 5+3=8
    a₂ = S₂ - S₁ = (20+6)-(5+3)=26-8=18
    d = 18-8=10
40. Solution:
    a = 4, d = 3
    100 = 4 + (n-1)×3
    96 = 3(n-1)
    n-1 = 32 → n = 33

Mixed Practice

41. Solution:
    S₅₀ = 50/2 (1+50) = 25×51 = 1275
42. Solution:
    a₁₀ = S₁₀ - S₉ = (200+30)-(162+27)=230-189=41
43. Solution:
    a = 20, d = -1.5
    20 + (n-1)(-1.5) < 0 20 - 1.5n + 1.5 < 0 21.5 < 1.5n → n > 14.33
    So, 15th term
44. If the 6th term of an AP is 22 and the 15th term is 49, find the common difference.
    Solution:
    a₆ = a + 5d = 22
    a₁₅ = a + 14d = 49
    14d - 5d = 49 - 22 → 9d = 27 → d = 3
45. Solution:
    S₅ = 40, S₁₀ = 130
    S₆-₁₀ = S₁₀ - S₅ = 90
    S₁₀ = 10/2 (a + a₁₀) = 5(a + a₁₀) = 130 → a + a₁₀ = 26
    Similarly, S₅ = 5/2 (2a + 4d) = 40 → 2a + 4d = 16 → a + 2d = 8
    a₁₀ = a + 9d
    a + a₁₀ = a + a + 9d = 2a + 9d = 26
    2a + 9d = 26
    2a + 4d = 16
    Subtract: 5d = 10 → d = 2
    a + 2d = 8 → a = 8 - 4 = 4
    AP: 4, 6, 8, 10, 12, …
46. Solution:
    b = (a + c)/2
    a + b + c = 18 → b = 18/3 = 6
47. Solution:
    Terms: a-d, a, a+d
    3a = 27 → a = 9
    (9-d)×9×(9+d) = 504
    (81 - d²)×9 = 504
    81 - d² = 56 → d² = 25 → d = 5
    Terms: 4, 9, 14
48. Solution:
    S₇ = 119, a = 6
    S₇ = 7/2 [2×6 + 6d] = (7/2)(12 + 6d) = 119
    12 + 6d = 34 → 6d = 22 → d = 11/3 ≈ 3.667
49. Solution:
    a = 15, d = -3
    a₂₀ = 15 + 19×(-3) = 15 - 57 = -42
50. Solution:
    a = 7, d = 6
    85 = 7 + (n-1)×6
    78 = 6(n-1)
    n-1 = 13 → n = 14

Challenging and Conceptual

51. Solution:
    a₁₀ = S₁₀ - S₉ = (300+10)-(243+9)=310-252=58
52. Solution:
    S₈ = 100, S₁₆ = 440
    S₉₋₁₆ = S₁₆ - S₈ = 340
    S₉₋₁₆ = 8/2 [2a + (2×8-1)d] = 4[2a + 15d] = 340 → 2a + 15d = 85
    S₈ = 4[2a + 7d] = 100 → 2a + 7d = 25
    Subtract: 8d = 60 → d = 7.5
    2a + 7×7.5 = 25 → 2a + 52.5 = 25 → 2a = -27.5 → a = -13.75
53. Solution:
    a = 100, d = -3
    100 + (n-1)(-3) < 0 100 - 3n + 3 < 0 103 < 3n → n > 34.33
    So, 35th term
54. Solution:
    a + 4d = 0
    a + 8d = 8
    Subtract: 4d = 8 → d = 2
    a = -8
    AP: -8, -6, -4, -2, 0, 2, 4, 6, 8, …
55. Solution:
    S₃₀ = 30/2 (a + 49) = 15(a + 49) = 750
    a + 49 = 50 → a = 1
56. Solution:
    Smallest odd: 1, Largest: 99, d = 2
    n = (99-1)/2 + 1 = 50
    S₅₀ = 50/2 (1+99) = 25×100 = 2500
57. Solution:
    a₁₁ = S₁₁ - S₁₀ = (484+11)-(400+10)=495-410=85
58. Solution:
    a + 6d = 20; a + 12d = 38
    6d = 18 → d = 3
    a = 20 - 18 = 2
    AP: 2, 5, 8, 11, …
59. Solution:
    a = 12, d = -3
    aₙ = 12 + (n-1)(-3) = -33
    -33 = 12 - 3n + 3
    -33 = 15 - 3n
    -48 = -3n → n = 16
60. Solution:
    S₁₆ = 8[2×4 + 15d] = 8[8 + 15d] = 400
    8 + 15d = 50 → 15d = 42 → d = 2.8

Real-World and Word Problems

61. Solution:
    Let first installment = a, d = 50, n = 10
    S₁₀ = 10/2 (2a + 9×50) = 5(2a + 450) = 5000
    2a + 450 = 1000 → 2a = 550 → a = 275
    Last installment: a₁₀ = 275 + 9×50 = 275 + 450 = 725
62. Solution:
    a = 20, d = 4, n = 15
    a₁₅ = 20 + 14×4 = 20 + 56 = 76
63. Solution:
    n = 12, S₁₂ = 1200, d = 10
    S₁₂ = 6[2a + 11×10] = 6[2a + 110] = 1200
    2a + 110 = 200 → 2a = 90 → a = 45
    Last share: a₁₂ = 45 + 11×10 = 45 + 110 = 155
64. Solution:
    a = 50, d = 10, n = 12
    a₁₂ = 50 + 11×10 = 50 + 110 = 160
65. Solution:
    a = 2000, d = 500, n = 10
    a₁₀ = 2000 + 9×500 = 2000 + 4500 = 6500
66. Solution:
    Number of gaps = 19
    Distance = 19×30 = 570 cm
67. Solution:
    Annual increase = 2% of 5000 = 100
    After 10 years: 5000 + 9×100 = 5900
    (Note: This is a linear approximation; actual population using GP would be higher.)
68. Solution:
    a = 10, d = 2, n = 25
    S₂₅ = 25/2 [2×10 + 24×2] = 12.5[20 + 48]=12.5×68=850
69. Solution:
    a = 8, d = 2, n = 30
    S₃₀ = 15[2×8 + 29×2]=15[16 + 58]=15×74=1110
70. Solution:
    a = 1, d = 0.5, n = 20
    a₂₀ = 1 + 19×0.5 = 1 + 9.5 = 10.5 km

Bottom Line: Regular practice with a variety of AP problems, from basic to advanced, is the best way to master this topic and build confidence for any exam.

Solving arithmetic progression questions can sometimes be tricky, especially if you overlook small details or rush through calculations. Here are some of the most frequent mistakes students make, along with advice on how to avoid them:

1. Confusing the Position of the Term (n) with the Term’s Value
   Many students mix up the position of a term (n) with the actual value of the term (aₙ). Remember, n refers to the place in the sequence, not the number itself. For example, in the sequence 3, 5, 7, 9, the third term (n=3) is 7, not 3.
2. Using the Wrong Formula
   Arithmetic progressions have specific formulas for the nth term and the sum of n terms. Sometimes, students mistakenly use formulas from geometric or harmonic progressions, leading to incorrect answers. Always double-check that you are using:
   - nth term: aₙ = a + (n-1)d - Sum: Sₙ = n/2 [2a + (n-1)d] or Sₙ = n/2 (a₁ + aₙ)
3. Incorrectly Identifying the Common Difference (d)
   Students sometimes calculate the common difference incorrectly, especially if the sequence includes negative numbers or fractions. Always subtract the previous term from the next term (d = a₂ - a₁), and check for consistency across several pairs of terms.
4. Ignoring Negative or Zero Common Differences
   Some students assume the common difference must be positive, but APs can decrease (d < 0) or remain constant (d = 0). Make sure to recognize and correctly handle these cases.
5. Misapplying the Sum Formula
   When using the sum formula, students sometimes substitute the wrong value for n, or use the term’s value instead of its position. Always ensure that n is the number of terms, not the value of the last term.
6. Forgetting to Check If a Sequence Is Actually an AP
   Before applying AP formulas, always check that the sequence has a constant difference between terms. Applying AP formulas to a non-AP sequence will lead to incorrect results.
7. Skipping Steps or Not Showing Working
   In a rush, students may skip steps or fail to write out their calculations clearly. This not only makes it harder to spot mistakes, but can also cost marks in exams. Always show your steps, especially when solving for unknowns.
8. Calculation and Arithmetic Errors
   Small mistakes in addition, subtraction, or multiplication can lead to incorrect answers. Always double-check your arithmetic, especially when working with negative numbers or fractions.
9. Incorrectly Finding Missing Terms
   When asked to find missing terms in a sequence, students sometimes forget to use the common difference consistently or to check their answers by substituting back into the sequence.
10. Not Answering What the Question Asks
    Sometimes, students find the nth term or sum, but forget to answer the specific question (e.g., “Which term is equal to 0?” or “How many terms are needed for the sum to reach 100?”). Always reread the question and make sure your final answer addresses what is being asked.

How to Avoid These Mistakes:

• Carefully read and understand the problem before starting.
• Write down all known values and formulas before substituting numbers.
• Check for a constant common difference before applying AP formulas.
• Show all your working and double-check calculations.
• Review your final answer to ensure it matches the question’s requirements.

Quick Note: By being aware of these common pitfalls and practicing step-by-step problem solving, you can significantly improve your accuracy and confidence in solving arithmetic progression questions.

Arithmetic progression is not just theoretical it appears in many practical situations. Understanding arithmetic and geometric progression questions can help you solve problems in finance, planning, and science.

• Financial Planning: Calculating savings with regular deposits, loan repayments (EMIs), and salary increments.
• Resource Distribution: Dividing prizes, scholarships, or resources in increasing or decreasing order.
• Scheduling: Setting up timetables or work shifts with regular intervals.
• Architecture and Design: Arranging seats in a theater, steps in a staircase, or tiles in patterns.
• Science and Nature: Counting patterns in petals, shells, or layers in geological formations.

Quick recap: Recognizing AP in real life helps you connect classroom math to everyday situations and practical problem-solving.

• Visualize the Sequence: Drawing a number line or sequence chart can help clarify the pattern.
• Work Both Forwards and Backwards: When finding missing terms, use the property of equal differences in both directions.
• Use Algebraic Manipulation: For complex problems, set up equations for unknowns and solve systematically.
• Practice Word Problems: Translate real-life scenarios into AP format for better understanding.
• Double-Check Calculations: Small mistakes in arithmetic can lead to wrong answers; always review your work.

Bottom Line: Mastering these strategies will help you tackle even the most challenging arithmetic progression MCQ and written questions.

Arithmetic progression questions are a cornerstone of mathematics, appearing in exams, competitions, and real-world situations. By understanding the key concepts, memorizing essential formulas, practicing a wide variety of problems, and learning from common mistakes, you can approach any AP question with confidence. Remember, consistent practice and stepwise problem-solving are the keys to success. Use the examples, practice problems, and tips in this guide to build your skills and excel in your mathematical journey.

Why It Matters

Arithmetic progression questions are a cornerstone of mathematics, appearing in exams, competitions, and real-world scenarios. Mastering AP not only boosts your exam scores but also sharpens your logical thinking and problem-solving skills for life.

Practical Advice for Learners

• Practice a variety of arithmetic progression questions regularly.
• Memorize key AP formulas and understand when to use them.
• Always check for a constant difference before applying AP concepts.
• Break down complex problems into smaller, manageable steps.
• Review mistakes to avoid repeating them.
• Apply AP concepts to real-world situations to deepen your understanding.

Q1: What is an arithmetic progression question?
A: It is a problem involving a sequence where each term increases or decreases by a fixed amount (the common difference). Arithmetic progression questions and arithmetic progression problems test your ability to use formulas and recognize patterns, and are often featured in arithmetic progression questions and solutions sections of textbooks and exams.

Q2: How do you solve AP questions step by step?
A: Identify the first term and common difference, choose the right formula, substitute values, solve for the unknown, and check your answer.

Q3: What is the most important formula in AP?
A: The nth term formula: ( a_n = a + (n-1)d ), and the sum formula: ( S_n = \frac{n}{2}[2a + (n-1)d] ). These are essential for solving arithmetic progression mcq and arithmetic progression quiz questions.

Q4: Why do AP questions appear so often in exams?
A: They test core mathematical skills: pattern recognition, logical reasoning, and formula application.

Q5: Can the common difference be negative or zero?
A: Yes. A negative ( d ) means the sequence decreases; zero means all terms are equal.

Q6: How can I check if a sequence is an AP?
A: Calculate the difference between consecutive terms. If the difference is always the same, it’s an AP. This check is crucial for arithmetic progression case study questions and arithmetic progression hard questions.