Q1 (Asked in 2024):

Find the 10th term of the AP: 2, 7, 12, …

Solution: a = 2, d = 7 – 2 = 5. a₁₀ = 2 + (10 – 1)5 = 2 + 45 = 47.

Formula Used: aₙ = a + (n – 1)d

Q2 (Asked in 2023):

Find the sum of the first 15 terms of the AP: 3, 8, 13, …

Solution: a = 3, d = 5. S₁₅ = 15/2 [2(3) + (15 – 1)5] = 15/2 (6 + 70) = 15 × 38 = 570.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q3 (Asked in 2022):

Which term of the AP: 5, 15, 25, … is 105?

Solution: a = 5, d = 10. aₙ = 105. 5 + (n – 1)10 = 105. (n – 1)10 = 100. n = 11.

Formula Used: aₙ = a + (n – 1)d

Q4 (Asked in 2021):

Find the common difference of the AP: 1/3, 5/3, 9/3, …

Solution: d = 5/3 – 1/3 = 4/3.

Formula Used: d = a₂ – a₁

Q5 (Asked in 2020):

The sum of the first n terms of an AP is 3n² + 5n. Find the first term and common difference.

Solution: S₁ = 3(1)² + 5(1) = 8 (first term, a = 8). S₂ = 3(2)² + 5(2) = 22. a₂ = S₂ – S₁ = 22 – 8 = 14. d = 14 – 8 = 6.

Formula Used: a = S₁, d = a₂ – a₁

Q6 (Asked in 2019):

Find the 20th term of the AP: -6, -3, 0, …

Solution: a = -6, d = -3 – (-6) = 3. a₂₀ = -6 + (20 – 1)3 = -6 + 57 = 51.

Formula Used: aₙ = a + (n – 1)d

Q7 (Asked in 2018):

Find the sum of the first 10 terms of the AP: 2, 5, 8, …

Solution: a = 2, d = 3. S₁₀ = 10/2 [2(2) + (10 – 1)3] = 5 (4 + 27) = 5 × 31 = 155.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q8 (Asked in 2017):

Is 184 a term of the AP: 3, 7, 11, …?

Solution: a = 3, d = 4. aₙ = 184. 3 + (n – 1)4 = 184. (n – 1)4 = 181. n = 181/4 + 1 = 46.25. Not an integer, so 184 is not a term.

Formula Used: aₙ = a + (n – 1)d

Q9 (Asked in 2016):

Find the number of terms in the AP: 7, 13, 19, …, 205.

Solution: a = 7, d = 6, l = 205. n = [(205 – 7)/6] + 1 = 33 + 1 = 34.

Formula Used: n = [(l – a)/d] + 1

Q10 (Asked in 2015):

The first term of an AP is 5, and the last term is 45. If the sum of the terms is 400, find the number of terms and the common difference.

Solution: a = 5, l = 45, Sₙ = 400. Sₙ = n/2 (a + l). 400 = n/2 (5 + 45). 400 = 25n. n = 16. aₙ = a + (n – 1)d. 45 = 5 + (16 – 1)d. 15d = 40. d = 8/3.

Formula Used: Sₙ = n/2 (a + l), aₙ = a + (n – 1)d

Q11 (Asked in 2014):

Find the 15th term of the AP: 10, 7, 4, …

Solution: a = 10, d = 7 – 10 = -3. a₁₅ = 10 + (15 – 1)(-3) = 10 – 42 = -32.

Formula Used: aₙ = a + (n – 1)d

Q12 (Asked in 2013):

Find the sum of the first 20 terms of the AP: 1, 4, 7, …

Solution: a = 1, d = 3. S₂₀ = 20/2 [2(1) + (20 – 1)3] = 10 (2 + 57) = 10 × 59 = 590.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q13 (Asked in 2012):

Which term of the AP: 21, 18, 15, … is 0?

Solution: a = 21, d = -3. aₙ = 0. 21 + (n – 1)(-3) = 0. 21 – 3n + 3 = 0. 3n = 24. n = 8.

Formula Used: aₙ = a + (n – 1)d

Q14 (Asked in 2011):

Find the common difference of the AP: -5, -1, 3, …

Solution: d = -1 – (-5) = 4.

Formula Used: d = a₂ – a₁

Q15 (Asked in 2010):

The sum of the first n terms of an AP is 4n – n². Find the first term and common difference.

Solution: S₁ = 4(1) – 1² = 3 (a = 3). S₂ = 4(2) – 2² = 4. a₂ = S₂ – S₁ = 4 – 3 = 1. d = 1 – 3 = -2.

Formula Used: a = S₁, d = a₂ – a₁

Q16 (Asked in 2024):

Find the 25th term of the AP: 3, 10, 17, …

Solution: a = 3, d = 7. a₂₅ = 3 + (25 – 1)7 = 3 + 168 = 171.

Formula Used: aₙ = a + (n – 1)d

Q17 (Asked in 2023):

Find the sum of the first 12 terms of the AP: 5, 9, 13, …

Solution: a = 5, d = 4. S₁₂ = 12/2 [2(5) + (12 – 1)4] = 6 (10 + 44) = 6 × 54 = 324.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q18 (Asked in 2022):

Which term of the AP: 8, 14, 20, … is 110?

Solution: a = 8, d = 6. aₙ = 110. 8 + (n – 1)6 = 110. (n – 1)6 = 102. n = 18.

Formula Used: aₙ = a + (n – 1)d

Q19 (Asked in 2021):

Find the number of terms in the AP: 9, 17, 25, …, 209.

Solution: a = 9, d = 8, l = 209. n = [(209 – 9)/8] + 1 = 25 + 1 = 26.

Formula Used: n = [(l – a)/d] + 1

Q20 (Asked in 2020):

The first term of an AP is 2, and the sum of the first 10 terms is 255. Find the common difference.

Solution: a = 2, S₁₀ = 255. S₁₀ = 10/2 [2(2) + (10 – 1)d] = 5 (4 + 9d) = 255. 4 + 9d = 51. 9d = 47. d = 47/9.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q21 (Asked in 2019):

Find the 30th term of the AP: 10, 7, 4, …

Solution: a = 10, d = -3. a₃₀ = 10 + (30 – 1)(-3) = 10 – 87 = -77.

Formula Used: aₙ = a + (n – 1)d

Q22 (Asked in 2018):

Find the sum of the first 25 terms of the AP: 2, 5, 8, …

Solution: a = 2, d = 3. S₂₅ = 25/2 [2(2) + (25 – 1)3] = 25/2 (4 + 72) = 25/2 × 76 = 950.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q23 (Asked in 2017):

Is 301 a term of the AP: 5, 11, 17, …?

Solution: a = 5, d = 6. aₙ = 301. 5 + (n – 1)6 = 301. (n – 1)6 = 296. n = 296/6 + 1 = 50.33. Not an integer, so 301 is not a term.

Formula Used: aₙ = a + (n – 1)d

Q24 (Asked in 2016):

Find the number of terms in the AP: 12, 18, 24, …, 300.

Solution: a = 12, d = 6, l = 300. n = [(300 – 12)/6] + 1 = 48 + 1 = 49.

Formula Used: n = [(l – a)/d] + 1

Q25 (Asked in 2015):

The first term of an AP is 3, and the 11th term is 35. Find the sum of the first 11 terms.

Solution: a = 3, a₁₁ = 35. a₁₁ = 3 + (11 – 1)d = 35. 3 + 10d = 35. 10d = 32. d = 3.2. S₁₁ = 11/2 (3 + 35) = 11/2 × 38 = 209.

Formula Used: aₙ = a + (n – 1)d, Sₙ = n/2 (a + l)

Q26 (Asked in 2014):

Find the 12th term of the AP: 4, 9, 14, …

Solution: a = 4, d = 5. a₁₂ = 4 + (12 – 1)5 = 4 + 55 = 59.

Formula Used: aₙ = a + (n – 1)d

Q27 (Asked in 2013):

Find the sum of the first 18 terms of the AP: 7, 10, 13, …

Solution: a = 7, d = 3. S₁₈ = 18/2 [2(7) + (18 – 1)3] = 9 (14 + 51) = 9 × 65 = 585.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q28 (Asked in 2012):

Which term of the AP: 20, 17, 14, … is -10?

Solution: a = 20, d = -3. aₙ = -10. 20 + (n – 1)(-3) = -10. 20 – 3n + 3 = -10. 3n = 33. n = 11.

Formula Used: aₙ = a + (n – 1)d

Q29 (Asked in 2011):

Find the common difference of the AP: 2, 8/3, 10/3, …

Solution: d = 8/3 – 2 = 2/3.

Formula Used: d = a₂ – a₁

Q30 (Asked in 2010):

The sum of the first n terms of an AP is 5n² + 3n. Find the first term and common difference.

Solution: S₁ = 5(1)² + 3(1) = 8 (a = 8). S₂ = 5(2)² + 3(2) = 26. a₂ = 26 – 8 = 18. d = 18 – 8 = 10.

Formula Used: a = S₁, d = a₂ – a₁

Q31 (Asked in 2024):

Find the 18th term of the AP: 1, 6, 11, …

Solution: a = 1, d = 5. a₁₈ = 1 + (18 – 1)5 = 1 + 85 = 86.

Formula Used: aₙ = a + (n – 1)d

Q32 (Asked in 2023):

Find the sum of the first 22 terms of the AP: 8, 3, -2, …

Solution: a = 8, d = -5. S₂₂ = 22/2 [2(8) + (22 – 1)(-5)] = 11 (16 – 105) = 11 × (-89) = -979.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q33 (Asked in 2022):

Which term of the AP: 3, 10, 17, … is 136?

Solution: a = 3, d = 7. aₙ = 136. 3 + (n – 1)7 = 136. (n – 1)7 = 133. n = 20.

Formula Used: aₙ = a + (n – 1)d

Q34 (Asked in 2021):

Find the number of terms in the AP: 5, 11, 17, …, 287.

Solution: a = 5, d = 6, l = 287. n = [(287 – 5)/6] + 1 = 47 + 1 = 48.

Formula Used: n = [(l – a)/d] + 1

Q35 (Asked in 2020):

The first term of an AP is 4, and the sum of the first 15 terms is 450. Find the common difference.

Solution: a = 4, S₁₅ = 450. S₁₅ = 15/2 [2(4) + (15 – 1)d] = 15/2 (8 + 14d) = 450. 8 + 14d = 60. 14d = 52. d = 26/7.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q36 (Asked in 2019):

Find the 16th term of the AP: 15, 12, 9, …

Solution: a = 15, d = -3. a₁₆ = 15 + (16 – 1)(-3) = 15 – 45 = -30.

Formula Used: aₙ = a + (n – 1)d

Q37 (Asked in 2018):

Find the sum of the first 30 terms of the AP: 1, 3, 5, …

Solution: a = 1, d = 2. S₃₀ = 30/2 [2(1) + (30 – 1)2] = 15 (2 + 58) = 15 × 60 = 900.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q38 (Asked in 2017):

Is 250 a term of the AP: 2, 7, 12, …?

Solution: a = 2, d = 5. aₙ = 250. 2 + (n – 1)5 = 250. (n – 1)5 = 248. n = 49.6. Not an integer, so 250 is not a term.

Formula Used: aₙ = a + (n – 1)d

Q39 (Asked in 2016):

Find the number of terms in the AP: 10, 16, 22, …, 310.

Solution: a = 10, d = 6, l = 310. n = [(310 – 10)/6] + 1 = 50 + 1 = 51.

Formula Used: n = [(l – a)/d] + 1

Q40 (Asked in 2015):

The first term of an AP is 6, and the 10th term is 51. Find the sum of the first 10 terms.

Solution: a = 6, a₁₀ = 51. a₁₀ = 6 + (10 – 1)d = 51. 6 + 9d = 51. 9d = 45. d = 5. S₁₀ = 10/2 (6 + 51) = 5 × 57 = 285.

Formula Used: aₙ = a + (n – 1)d, Sₙ = n/2 (a + l)

Q41 (Asked in 2014):

Find the 22nd term of the AP: 7, 11, 15, …

Solution: a = 7, d = 4. a₂₂ = 7 + (22 – 1)4 = 7 + 84 = 91.

Formula Used: aₙ = a + (n – 1)d

Q42 (Asked in 2013):

Find the sum of the first 16 terms of the AP: 4, 8, 12, …

Solution: a = 4, d = 4. S₁₆ = 16/2 [2(4) + (16 – 1)4] = 8 (8 + 60) = 8 × 68 = 544.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q43 (Asked in 2012):

Which term of the AP: 25, 22, 19, … is -2?

Solution: a = 25, d = -3. aₙ = -2. 25 + (n – 1)(-3) = -2. 25 – 3n + 3 = -2. 3n = 30. n = 10.

Formula Used: aₙ = a + (n – 1)d

Q44 (Asked in 2011):

Find the common difference of the AP: -2, 1, 4, …

Solution: d = 1 – (-2) = 3.

Formula Used: d = a₂ – a₁

Q45 (Asked in 2010):

The sum of the first n terms of an AP is 2n² + n. Find the first term and common difference.

Solution: S₁ = 2(1)² + 1 = 3 (a = 3). S₂ = 2(2)² + 2 = 10. a₂ = 10 – 3 = 7. d = 7 – 3 = 4.

Formula Used: a = S₁, d = a₂ – a₁

Q46 (Asked in 2024):

Find the 14th term of the AP: 6, 11, 16, …

Solution: a = 6, d = 5. a₁₄ = 6 + (14 – 1)5 = 6 + 65 = 71.

Formula Used: aₙ = a + (n – 1)d

Q47 (Asked in 2023):

Find the sum of the first 20 terms of the AP: 9, 7, 5, …

Solution: a = 9, d = -2. S₂₀ = 20/2 [2(9) + (20 – 1)(-2)] = 10 (18 – 38) = 10 × (-20) = -200.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q48 (Asked in 2022):

Which term of the AP: 4, 9, 14, … is 124?

Solution: a = 4, d = 5. aₙ = 124. 4 + (n – 1)5 = 124. (n – 1)5 = 120. n = 25.

Formula Used: aₙ = a + (n – 1)d

Q49 (Asked in 2021):

Find the number of terms in the AP: 8, 14, 20, …, 272.

Solution: a = 8, d = 6, l = 272. n = [(272 – 8)/6] + 1 = 44 + 1 = 45.

Formula Used: n = [(l – a)/d] + 1

Q50 (Asked in 2020):

The first term of an AP is 7, and the sum of the first 12 terms is 258. Find the common difference.

Solution: a = 7, S₁₂ = 258. S₁₂ = 12/2 [2(7) + (12 – 1)d] = 6 (14 + 11d) = 258. 14 + 11d = 43. 11d = 29. d = 29/11.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q51 (Asked in 2019):

Find the 17th term of the AP: 2, 8, 14, …

Solution: a = 2, d = 6. a₁₇ = 2 + (17 – 1)6 = 2 + 96 = 98.

Formula Used: aₙ = a + (n – 1)d

Q52 (Asked in 2018):

Find the sum of the first 15 terms of the AP: 10, 6, 2, …

Solution: a = 10, d = -4. S₁₅ = 15/2 [2(10) + (15 – 1)(-4)] = 15/2 (20 – 56) = 15/2 × (-36) = -270.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q53 (Asked in 2017):

Is 200 a term of the AP: 3, 9, 15, …?

Solution: a = 3, d = 6. aₙ = 200. 3 + (n – 1)6 = 200. (n – 1)6 = 197. n = 33.83. Not an integer, so 200 is not a term.

Formula Used: aₙ = a + (n – 1)d

Q54 (Asked in 2016):

Find the number of terms in the AP: 7, 13, 19, …, 247.

Solution: a = 7, d = 6, l = 247. n = [(247 – 7)/6] + 1 = 40 + 1 = 41.

Formula Used: n = [(l – a)/d] + 1

Q55 (Asked in 2015):

The first term of an AP is 2, and the 8th term is 23. Find the sum of the first 8 terms.

Solution: a = 2, a₈ = 23. a₈ = 2 + (8 – 1)d = 23. 2 + 7d = 23. 7d = 21. d = 3. S₈ = 8/2 (2 + 23) = 4 × 25 = 100.

Formula Used: aₙ = a + (n – 1)d, Sₙ = n/2 (a + l)

Q56 (Asked in 2014):

Find the 20th term of the AP: 9, 14, 19, …

Solution: a = 9, d = 5. a₂₀ = 9 + (20 – 1)5 = 9 + 95 = 104.

Formula Used: aₙ = a + (n – 1)d

Q57 (Asked in 2013):

Find the sum of the first 24 terms of the AP: 5, 8, 11, …

Solution: a = 5, d = 3. S₂₄ = 24/2 [2(5) + (24 – 1)3] = 12 (10 + 69) = 12 × 79 = 948.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q58 (Asked in 2012):

Which term of the AP: 30, 27, 24, … is 0?

Solution: a = 30, d = -3. aₙ = 0. 30 + (n – 1)(-3) = 0. 30 – 3n + 3 = 0. 3n = 33. n = 11.

Formula Used: aₙ = a + (n – 1)d

Q59 (Asked in 2011):

Find the common difference of the AP: -10, -7, -4, …

Solution: d = -7 – (-10) = 3.

Formula Used: d = a₂ – a₁

Q60 (Asked in 2010):

The sum of the first n terms of an AP is 3n² + 2n. Find the first term and common difference.

Solution: S₁ = 3(1)² + 2(1) = 5 (a = 5). S₂ = 3(2)² + 2(2) = 16. a₂ = 16 – 5 = 11. d = 11 – 5 = 6.

Formula Used: a = S₁, d = a₂ – a₁

Q61 (Asked in 2024):

Find the 19th term of the AP: 4, 10, 16, …

Solution: a = 4, d = 6. a₁₉ = 4 + (19 – 1)6 = 4 + 108 = 112.

Formula Used: aₙ = a + (n – 1)d

Q62 (Asked in 2023):

Find the sum of the first 14 terms of the AP: 6, 9, 12, …

Solution: a = 6, d = 3. S₁₄ = 14/2 [2(6) + (14 – 1)3] = 7 (12 + 39) = 7 × 51 = 357.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q63 (Asked in 2022):

Which term of the AP: 7, 13, 19, … is 151?

Solution: a = 7, d = 6. aₙ = 151. 7 + (n – 1)6 = 151. (n – 1)6 = 144. n = 25.

Formula Used: aₙ = a + (n – 1)d

Q64 (Asked in 2021):

Find the number of terms in the AP: 3, 8, 13, …, 253.

Solution: a = 3, d = 5, l = 253. n = [(253 – 3)/5] + 1 = 50 + 1 = 51.

Formula Used: n = [(l – a)/d] + 1

Q65 (Asked in 2020):

The first term of an AP is 5, and the sum of the first 20 terms is 590. Find the common difference.

Solution: a = 5, S₂₀ = 590. S₂₀ = 20/2 [2(5) + (20 – 1)d] = 10 (10 + 19d) = 590. 10 + 19d = 59. 19d = 49. d = 49/19.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q66 (Asked in 2019):

Find the 18th term of the AP: 12, 9, 6, …

Solution: a = 12, d = -3. a₁₈ = 12 + (18 – 1)(-3) = 12 – 51 = -39.

Formula Used: aₙ = a + (n – 1)d

Q67 (Asked in 2018):

Find the sum of the first 28 terms of the AP: 2, 6, 10, …

Solution: a = 2, d = 4. S₂₈ = 28/2 [2(2) + (28 – 1)4] = 14 (4 + 108) = 14 × 112 = 1568.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q68 (Asked in 2017):

Is 180 a term of the AP: 4, 10, 16, …?

Solution: a = 4, d = 6. aₙ = 180. 4 + (n – 1)6 = 180. (n – 1)6 = 176. n = 29.33. Not an integer, so 180 is not a term.

Formula Used: aₙ = a + (n – 1)d

Q69 (Asked in 2016):

Find the number of terms in the AP: 5, 11, 17, …, 299.

Solution: a = 5, d = 6, l = 299. n = [(299 – 5)/6] + 1 = 49 + 1 = 50.

Formula Used: n = [(l – a)/d] + 1

Q70 (Asked in 2015):

The first term of an AP is 3, and the 12th term is 36. Find the sum of the first 12 terms.

Solution: a = 3, a₁₂ = 36. a₁₂ = 3 + (12 – 1)d = 36. 3 + 11d = 36. 11d = 33. d = 3. S₁₂ = 12/2 (3 + 36) = 6 × 39 = 234.

Formula Used: aₙ = a + (n – 1)d, Sₙ = n/2 (a + l)

Q71 (Asked in 2014):

Find the 15th term of the AP: 8, 12, 16, …

Solution: a = 8, d = 4. a₁₅ = 8 + (15 – 1)4 = 8 + 56 = 64.

Formula Used: aₙ = a + (n – 1)d

Q72 (Asked in 2013):

Find the sum of the first 22 terms of the AP: 7, 10, 13, …

Solution: a = 7, d = 3. S₂₂ = 22/2 [2(7) + (22 – 1)3] = 11 (14 + 63) = 11 × 77 = 847.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q73 (Asked in 2012):

Which term of the AP: 27, 24, 21, … is -3?

Solution: a = 27, d = -3. aₙ = -3. 27 + (n – 1)(-3) = -3. 27 – 3n + 3 = -3. 3n = 33. n = 11.

Formula Used: aₙ = a + (n – 1)d

Q74 (Asked in 2011):

Find the common difference of the AP: -3, 0, 3, …

Solution: d = 0 – (-3) = 3.

Formula Used: d = a₂ – a₁

Q75 (Asked in 2010):

The sum of the first n terms of an AP is 4n² + 2n. Find the first term and common difference.

Solution: S₁ = 4(1)² + 2(1) = 6 (a = 6). S₂ = 4(2)² + 2(2) = 20. a₂ = 20 – 6 = 14. d = 14 – 6 = 8.

Formula Used: a = S₁, d = a₂ – a₁

Q76 (Asked in 2024):

Find the 21st term of the AP: 5, 11, 17, …

Solution: a = 5, d = 6. a₂₁ = 5 + (21 – 1)6 = 5 + 120 = 125.

Formula Used: aₙ = a + (n – 1)d

Q77 (Asked in 2023):

Find the sum of the first 18 terms of the AP: 4, 7, 10, …

Solution: a = 4, d = 3. S₁₈ = 18/2 [2(4) + (18 – 1)3] = 9 (8 + 51) = 9 × 59 = 531.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q78 (Asked in 2022):

Which term of the AP: 6, 12, 18, … is 186?

Solution: a = 6, d = 6. aₙ = 186. 6 + (n – 1)6 = 186. (n – 1)6 = 180. n = 31.

Formula Used: aₙ = a + (n – 1)d

Q79 (Asked in 2021):

Find the number of terms in the AP: 2, 9, 16, …, 303.

Solution: a = 2, d = 7, l = 303. n = [(303 – 2)/7] + 1 = 43 + 1 = 44.

Formula Used: n = [(l – a)/d] + 1

Q80 (Asked in 2020):

The first term of an AP is 8, and the sum of the first 15 terms is 360. Find the common difference.

Solution: a = 8, S₁₅ = 360. S₁₅ = 15/2 [2(8) + (15 – 1)d] = 15/2 (16 + 14d) = 360. 16 + 14d = 48. 14d = 32. d = 16/7.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q81 (Asked in 2019):

Find the 23rd term of the AP: 10, 7, 4, …

Solution: a = 10, d = -3. a₂₃ = 10 + (23 – 1)(-3) = 10 – 66 = -56.

Formula Used: aₙ = a + (n – 1)d

Q82 (Asked in 2018):

Find the sum of the first 20 terms of the AP: 3, 6, 9, …

Solution: a = 3, d = 3. S₂₀ = 20/2 [2(3) + (20 – 1)3] = 10 (6 + 57) = 10 × 63 = 630.

Formula Used: Sₙ = n/2 [2a + (n – 1)d]

Q83 (Asked in 2017):

Is 240 a term of the AP: 5, 10, 15, …?

Solution: a = 5, d = 5. aₙ = 240. 5 + (n – 1)5 = 240. (n – 1)5 = 235. n = 48. Integer, so 240 is a term.

Formula Used: aₙ = a + (n – 1)d

Q84 (Asked in 2016):

Find the number of terms in the AP: 6, 12, 18, …, 360.

Solution: a = 6, d = 6, l = 360. n = [(360 – 6)/6] + 1 = 59 + 1 = 60.

Formula Used: n = [(l – a)/d] + 1

Q85 (Asked in 2015):

The first term of an AP is 4, and the 15th term is 60. Find the sum of the first 15 terms.

Solution: a = 4, a₁₅ = 60. a₁₅ = 4 + (15 – 1)d = 60. 4 + 14d = 60. 14d = 56. d = 4. S₁₅ = 15/2 (4 + 60) = 15/2 ×

Shekhar

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