Class 10 Arithmetic Progressions: Practice and Worked Solutions

Worked example: one row versus all rows

A hall has 18 seats in its first row and 3 more seats in each next row. Find the seats in row 12 and the total seats in the first 12 rows. Here a = 18, d = 3 and n = 12.

For row 12 use aₙ = a + (n − 1)d: a₁₂ = 18 + 11 × 3 = 51 seats. There are 11 increases between the first and twelfth rows, not 12.

For the total use Sₙ = n/2 × [2a + (n − 1)d]: S₁₂ = 12/2 × [36 + 33] = 414 seats. Check using the first and last rows: 12 × (18 + 51)/2 = 414. The word total changes the quantity you are finding.

Read the quantity before choosing a formula

• Check successive differences. 5, 9, 13 is an AP because both differences are 4; 5, 10, 20 is not an AP because the differences change.
• Use aₙ = a + (n − 1)d for a particular term. Use the same equation and solve for n if you are asked where a value appears.
• Use Sₙ = n/2 × [2a + (n − 1)d] for a total. If the last term l is known, Sₙ = n(a + l)/2 gives the same result.
• Count terms inclusively. From the first term to the nth term there are n − 1 gaps. In a decreasing AP the common difference is negative.
• A real situation is an AP only when its increase or decrease is constant. Repeated percentage growth is not a constant difference.

Original practice: terms, totals and interpretation

For a short session attempt Q1, Q3 and Q7. Write a, d and the required quantity first. Give units for contexts and explain any rejected index. Read the solutions only after attempting the question; this is a practice set, not a scored diagnostic.

• Q1. For 7, 11, 15, … find d and the 20th term.
• Q2. Is 100 a term of 7, 11, 15, …? Justify your answer by solving for its position.
• Q3. Find the sum of the first 15 terms of 2, 5, 8, … .
• Q4. An AP has a₅ = 22 and a₉ = 38. Find a and d, then find a₁₂.
• Q5. For 30, 27, 24, … find the first negative term and its position.
• Q6. A student saves ₹40 in week 1 and increases the weekly saving by ₹10 each week. Find the saving in week 8 and the total saved over weeks 1–8.
• Q7. Case study: a display has 12 tiles in its first row and 2 more in every next row. (a) Find the number in row 10. (b) Find the total in the first 10 rows. (c) Can one row contain exactly 31 tiles? Explain.
• Q8. Find how many terms of 3, 6, 9, … have a sum of 165. Show why the accepted solution is a valid count.

Solutions: keep the term and total separate

Q1. d = 11 − 7 = 4. a₂₀ = 7 + 19 × 4 = 83. Using 20 × 4 would count one extra increase.

Q2. Set 7 + (n − 1)4 = 100. Then 4(n − 1) = 93, so n = 24.25. A term position is a positive integer, so 100 is not in this AP. The nearby terms are a₂₄ = 99 and a₂₅ = 103.

Q3. a = 2, d = 3, n = 15. S₁₅ = 15/2 × [4 + 14 × 3] = 15/2 × 46 = 345. Check that the last term is 44: 15 × (2 + 44)/2 = 345.

Q4. a + 4d = 22 and a + 8d = 38. Subtract to get 4d = 16, so d = 4. Then a = 6 and a₁₂ = 6 + 11 × 4 = 50. The four-position gap between terms 5 and 9 explains why the difference is divided by 4.

Q5. aₙ = 30 − 3(n − 1) = 33 − 3n. A negative term requires n > 11. The first is n = 12, with a₁₂ = −3. The 11th term is zero; zero is not negative.

Q6. a = 40, d = 10. Week 8 saving is a₈ = 40 + 7 × 10 = ₹110. Total saving is S₈ = 8/2 × [80 + 70] = ₹600. The weekly amount and accumulated amount must have separate labels.

Q7. a = 12 and d = 2. (a) a₁₀ = 12 + 9 × 2 = 30 tiles. (b) S₁₀ = 10/2 × [24 + 18] = 210 tiles. (c) 12 + 2(n − 1) = 31 gives n = 10.5, so no row has 31 tiles. Every row contains an even number, which provides another check.

Q8. a = 3, d = 3. Sₙ = n/2 × [6 + 3(n − 1)] = 3n(n + 1)/2. Set this equal to 165: n² + n − 110 = (n + 11)(n − 10) = 0. Accept n = 10 and reject −11 because a count cannot be negative. The last term is 30; 10 × (3 + 30)/2 = 165.

Use the first wrong step to choose your next practice

• Off by one: write the first three terms and count the gaps. Rebuild a₄ as a + 3d before returning to Q1 or Q7.
• Using a term formula for a total: underline 'in week 8' and 'over weeks 1–8' in Q6. Draw eight boxes, label the individual amounts and explain what is being added.
• Wrong difference: subtract the two known terms in Q4 and divide by the difference in their positions. Check the resulting a and d in both given equations.
• Accepting an impossible position: use Q2 and Q7 to explain why rounding n creates an answer to a different question. Verify the two surrounding terms instead.

Check the correction on changed questions

Retest after a break: R1. Find a₁₀ and S₁₀ for 5, 8, 11, … . R2. A hall has 14 seats in row 1 and 4 extra seats per subsequent row. Find the first row with more than 50 seats, and the total seats up to that row.

Retest checks: R1 has a₁₀ = 5 + 9 × 3 = 32 and S₁₀ = 10(5 + 32)/2 = 185. R2 requires 14 + 4(n − 1) > 50, so n > 10. Row 11 is first, with 54 seats. The total is 11(14 + 54)/2 = 374; row 10 has exactly 50.

Show a parent or tutor which quantity you chose, where the first mistake appeared and how the retest changed. If you can complete the set, return to your official paper; do not infer a full syllabus score from a small chapter exercise.

Connect AP work to algebra and the official paper

Q8 combines a sum with a quadratic equation. If finding or interpreting its roots was the difficult part, use the quadratic-equation practice next. Then attempt the relevant questions in your official Basic or Standard sample paper and check the written steps with its marking scheme.