Word Problems on Arithmetic Progressions
Word problems on Arithmetic Progressions (AP) are application-based questions from Chapter 5 of the NCERT Class 10 Mathematics textbook.
These problems require students to:
- Identify that a given situation involves an AP.
- Use the nth term formula or the sum formula to form equations.
- Solve the equations to find the unknown (number of terms, common difference, a specific term, or the sum).
AP word problems commonly appear in contexts like savings, production, rows of seats, stacking, distance covered in successive intervals, and salary increments.
What is Word Problems on Arithmetic Progressions — Types, Formulas & Solved Examples?
Definition: An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This constant is called the common difference (d).
General form: a, a + d, a + 2d, a + 3d, ...
where:
- a = first term
- d = common difference
- n = number of terms
Key properties:
- If d > 0, the AP is increasing.
- If d < 0, the AP is decreasing.
- If d = 0, all terms are equal.
Word Problems on Arithmetic Progressions Formula
Formulas for Arithmetic Progressions:
nth term (general term):
aₙ = a + (n − 1)d
Sum of first n terms:
Sₙ = n/2 [2a + (n − 1)d]
or equivalently
Sₙ = n/2 (a + l)
where l = last term = a + (n − 1)d.
Useful derived formulas:
| To Find | Formula |
|---|---|
| Common difference | d = (aₙ − a) / (n − 1) |
| Number of terms | n = (aₙ − a) / d + 1 |
| nth term from the end | l − (n − 1)d |
| Sum when last term is known | Sₙ = n(a + l)/2 |
Derivation and Proof
Approach to solving AP word problems:
- Read carefully. Identify whether the problem involves a sequence with a constant difference.
- Determine the AP parameters: What is a (first term)? What is d (common difference)?
- Decide which formula to use:
- If the problem asks for a specific term, use aₙ = a + (n − 1)d.
- If the problem asks for a sum, use Sₙ = n/2[2a + (n − 1)d].
- If the problem involves three consecutive terms, let them be a − d, a, a + d.
- Form the equation using the given condition.
- Solve for the unknown.
- Verify the answer.
Convenient substitutions for AP terms:
| Number of Terms | Let the Terms Be |
|---|---|
| 3 terms | a − d, a, a + d |
| 4 terms | a − 3d, a − d, a + d, a + 3d |
| 5 terms | a − 2d, a − d, a, a + d, a + 2d |
These substitutions simplify the sum (common difference terms cancel).
Types and Properties
Common types of AP word problems:
Type 1: Savings and deposits
- Saving increasing amounts each month/week.
- Example: Saved ₹100 in month 1, ₹150 in month 2, ₹200 in month 3.
Type 2: Production and manufacturing
- Output increasing by a fixed amount each day/year.
Type 3: Rows, seats, and stacking
- Number of items in successive rows forming an AP.
Type 4: Distance and motion
- Distance covered in successive seconds/minutes forms an AP.
Type 5: Salary increments
- Annual salary with fixed increment.
Type 6: Sum of natural numbers, multiples
- Find the sum of first n multiples of a given number.
Type 7: Finding unknown terms or common difference
- Given the sum and the number of terms, find a or d.
Methods
Method 1: Using nth term formula
- When the problem asks for a specific term, or when the last term is given.
- Formula: aₙ = a + (n − 1)d
- Use when: "In which year will...", "find the 15th term", "what is the last term"
Method 2: Using sum formula
- When the problem asks for the total/sum of terms.
- Formula: Sₙ = n/2 [2a + (n − 1)d]
- Alternative: Sₙ = n/2 (a + l) when the last term l is known.
- Use when: "find total savings", "total production", "sum of all multiples"
Method 3: Assuming symmetric terms
- When 3 or 5 terms are in AP and their sum is given.
- For 3 terms: let them be a − d, a, a + d. Sum = 3a (the d terms cancel).
- For 4 terms: let them be a − 3d, a − d, a + d, a + 3d. Sum = 4a.
- For 5 terms: let them be a − 2d, a − d, a, a + d, a + 2d. Sum = 5a.
- This technique simplifies the algebra greatly when the sum is given.
Method 4: Combining nth term and sum
- Some problems give two conditions (e.g., a₃ = 600 and a₇ = 700). This gives two equations in a and d.
- Solve simultaneously to find a and d, then compute what is asked.
Method 5: Finding sum of multiples between limits
- Find the first multiple above the lower limit (use ceiling of lower/k, then multiply by k).
- Find the last multiple below the upper limit (use floor of upper/k, then multiply by k).
- These form an AP with common difference k.
- Find n using aₙ = a + (n−1)d, then compute sum using Sₙ = n/2(a + l).
Common mistakes:
- Confusing the nth term formula with the sum formula — using aₙ when the problem asks for Sₙ, or vice versa.
- Forgetting that n must be a positive integer — rejecting fractional or negative values.
- Using the wrong value of a or d from the context — e.g., treating the 3rd year production as the first term.
- Forgetting to check both roots of the quadratic when the sum formula gives n² — one root is usually negative and must be rejected.
- Not verifying the answer — always check by computing the actual sum or term.
Solved Examples
Example 1: Seats in a Flower Bed
Problem: In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there, and how many rose plants are there in total?
Solution:
Given:
- a = 23, d = 21 − 23 = −2, last term l = 5
Finding n:
- aₙ = a + (n − 1)d
- 5 = 23 + (n − 1)(−2)
- 5 = 25 − 2n
- 2n = 20, so n = 10
Finding total:
- Sₙ = n/2(a + l) = 10/2(23 + 5) = 5 × 28 = 140
Verification: a₁₀ = 23 + 9(−2) = 23 − 18 = 5 ✓
Answer: There are 10 rows and 140 rose plants in total.
Example 2: Sum of Multiples
Problem: Find the sum of all multiples of 7 lying between 100 and 300.
Solution:
Given:
- First multiple of 7 > 100: 7 × 15 = 105
- Last multiple of 7 < 300: 7 × 42 = 294
AP: a = 105, d = 7, l = 294
Finding n:
- 294 = 105 + (n − 1)(7)
- 189 = 7(n − 1)
- n − 1 = 27, so n = 28
Finding sum:
- Sₙ = n/2(a + l) = 28/2(105 + 294) = 14 × 399 = 5586
Answer: The sum is 5586.
Example 3: Salary Increment
Problem: Subba Rao started with a salary of ₹5,000 per month and receives an annual increment of ₹200. In which year will his salary reach ₹7,000?
Solution:
Given:
- a = 5000, d = 200, aₙ = 7000
Using: aₙ = a + (n − 1)d
- 7000 = 5000 + (n − 1)(200)
- 2000 = 200(n − 1)
- n − 1 = 10, so n = 11
Answer: His salary will reach ₹7,000 in the 11th year.
Example 4: Production Problem
Problem: A manufacturer produces 600 TV sets in the third year and 700 in the seventh year. Assuming production increases by a fixed number each year, find the production in the first year and total production in 7 years.
Solution:
Given:
- a₃ = 600, a₇ = 700
Finding a and d:
- a + 2d = 600 ... (i)
- a + 6d = 700 ... (ii)
- Subtract (i) from (ii): 4d = 100 → d = 25
- From (i): a + 50 = 600 → a = 550
Total in 7 years:
- S₇ = 7/2[2(550) + 6(25)] = 7/2[1100 + 150] = 7/2 × 1250 = 4375
Verification: a₃ = 550 + 50 = 600 ✓, a₇ = 550 + 150 = 700 ✓
Answer: First year production = 550. Total in 7 years = 4375.
Example 5: Stacking Problem — Logs
Problem: 200 logs are stacked in rows. The bottom row has 20 logs, the next has 19, and so on. In how many rows are the 200 logs placed, and how many logs are in the top row?
Solution:
Given:
- a = 20, d = −1, Sₙ = 200
Using: Sₙ = n/2[2a + (n − 1)d]
- 200 = n/2[40 + (n − 1)(−1)]
- 200 = n/2[41 − n]
- 400 = 41n − n²
- n² − 41n + 400 = 0
- (n − 16)(n − 25) = 0
- n = 16 or n = 25
Check n = 25: a₂₅ = 20 + 24(−1) = −4 (negative, impossible).
So n = 16. Top row: a₁₆ = 20 + 15(−1) = 5.
Verification: S₁₆ = 16/2(20 + 5) = 8 × 25 = 200 ✓
Answer: 16 rows, with 5 logs in the top row.
Example 6: Distance Covered — Motion
Problem: A body covers 5 m in the first second, 15 m in the second, 25 m in the third, and so on. Find the total distance covered in 8 seconds.
Solution:
Given:
- a = 5, d = 10, n = 8
Using: Sₙ = n/2[2a + (n − 1)d]
- S₈ = 8/2[2(5) + 7(10)] = 4[10 + 70] = 4 × 80 = 320
Answer: Total distance = 320 m.
Example 7: Finding Three Terms in AP
Problem: The sum of three numbers in AP is 27 and their product is 648. Find the numbers.
Solution:
Given:
- Sum = 27, Product = 648
Steps:
- Let the three numbers be a − d, a, a + d.
- Sum: 3a = 27 → a = 9
- Product: (9 − d)(9)(9 + d) = 648
- 9(81 − d²) = 648
- 81 − d² = 72
- d² = 9 → d = 3 or d = −3
If d = 3: Numbers are 6, 9, 12.
Verification: 6 + 9 + 12 = 27 ✓, 6 × 9 × 12 = 648 ✓
Answer: The numbers are 6, 9, and 12.
Example 8: Finding Number of Terms for a Given Sum
Problem: How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?
Solution:
Given:
- a = 9, d = 8, Sₙ = 636
Using: Sₙ = n/2[2a + (n − 1)d]
- 636 = n/2[18 + 8(n − 1)]
- 636 = n/2[10 + 8n]
- 1272 = 10n + 8n²
- 8n² + 10n − 1272 = 0
- 4n² + 5n − 636 = 0
- n = [−5 ± √(25 + 10176)] / 8 = [−5 ± √10201] / 8
- √10201 = 101
- n = (−5 + 101)/8 = 96/8 = 12
Verification: S₁₂ = 12/2[18 + 88] = 6 × 106 = 636 ✓
Answer: 12 terms are needed.
Example 9: Penalty Problem
Problem: A contract specifies a penalty for delay: ₹200 for the first day, ₹250 for the second, ₹300 for the third, and so on. The total penalty is ₹27,750. How many days was the project delayed?
Solution:
Given:
- a = 200, d = 50, Sₙ = 27750
Using: Sₙ = n/2[2a + (n − 1)d]
- 27750 = n/2[400 + 50(n − 1)]
- 27750 = n/2[350 + 50n]
- 55500 = 350n + 50n²
- 50n² + 350n − 55500 = 0
- n² + 7n − 1110 = 0
- (n + 37)(n − 30) = 0
- n = 30 (rejecting −37)
Verification: S₃₀ = 30/2[400 + 1450] = 15 × 1850 = 27750 ✓
Answer: The project was delayed by 30 days.
Example 10: Total Earnings Over Years
Problem: A man starts saving ₹100 in January and increases his savings by ₹50 every month. What are his total savings at the end of 12 months?
Solution:
Given:
- a = 100, d = 50, n = 12
Using: Sₙ = n/2[2a + (n − 1)d]
- S₁₂ = 12/2[200 + 11(50)]
- = 6[200 + 550]
- = 6 × 750 = 4500
Verification: Last month savings = 100 + 11(50) = 650. S₁₂ = 12/2(100 + 650) = 6 × 750 = 4500 ✓
Answer: Total savings = ₹4,500.
Real-World Applications
Real-life applications of AP word problems:
- Financial planning — savings schemes with fixed monthly increments.
- Salary structures — calculating total earnings with annual fixed increments.
- Construction — counting bricks, tiles, or logs in tapered stacks.
- Seating arrangements — stadium or auditorium seating with increasing seats per row.
- Physics — distance covered under uniform acceleration in successive intervals.
- Manufacturing — production targets increasing by a fixed amount.
- Penalty calculations — progressive penalties for delay.
- Sports — training schedules with increasing repetitions.
Key Points to Remember
- An AP has first term a and common difference d. The nth term is aₙ = a + (n−1)d.
- Sum of first n terms: Sₙ = n/2[2a + (n−1)d] or Sₙ = n/2(a + l).
- For word problems, identify a, d, and what is being asked (n, aₙ, or Sₙ).
- When 3 terms are in AP, use a − d, a, a + d (sum = 3a).
- The sum formula gives a quadratic in n — only positive integer values of n are valid.
- If d is negative, the AP is decreasing — check that terms remain positive.
- Multiples of k between two numbers form an AP with d = k.
- Always verify by computing the term or sum and checking against the given condition.
- In CBSE exams, AP word problems carry 4–5 marks and require clear working.
- Know the formulas by heart — they are the foundation for all AP problems.
Practice Problems
- Find the sum of first 20 terms of the AP: 1, 4, 7, 10, ...
- How many terms of the AP 5, 11, 17, ... must be taken so that the sum is 670?
- The first term of an AP is 5, the last term is 45, and the sum is 400. Find the number of terms and common difference.
- Find the sum of all two-digit numbers divisible by 3.
- A sum of ₹700 is to be used to give 7 cash prizes. If each prize is ₹20 less than the preceding prize, find the value of each prize.
- In an AP, the sum of first 6 terms is 42 and the sum of first 10 terms is 150. Find the AP.
- A manufacturer produces 500 items in the first year and increases production by 50 each year. In how many years will total production reach 8000?
- The sum of 4th and 8th terms of an AP is 24. The sum of 6th and 10th terms is 44. Find the AP.
Frequently Asked Questions
Q1. How do I identify an AP in a word problem?
If a quantity changes by a fixed amount in each step (e.g., saves ₹5 more each week, produces 50 more each year, each row has 2 fewer seats), the situation involves an AP.
Q2. Which formula should I use — nth term or sum?
If the problem asks for a specific term, position, or the last term, use aₙ = a + (n−1)d. If it asks for the total or cumulative value, use Sₙ = n/2[2a + (n−1)d].
Q3. What if the sum formula gives a quadratic with two positive roots?
Check both roots against the context. If one root makes a term negative (e.g., negative number of logs), reject it. If both are valid, the problem may have two correct answers.
Q4. How do I find multiples of a number between two limits?
Find the first multiple above the lower limit and the last multiple below the upper limit. These form an AP with d = the number. Use the nth term formula to find n, then the sum formula.
Q5. What substitution should I use for 3 terms in AP?
Let the terms be a − d, a, a + d. Their sum is 3a (d cancels out), which simplifies calculation. For 4 terms, use a − 3d, a − d, a + d, a + 3d.
Q6. Can d be negative in word problems?
Yes. A decreasing AP has negative d. Examples: logs decreasing per row, savings decreasing each month. Ensure the terms remain non-negative in context.
Q7. How are AP word problems scored in CBSE exams?
AP word problems carry 4–5 marks. Marks are given for: identifying a and d (1 mark), writing the correct formula (1 mark), solving (1–2 marks), and the final answer (1 mark).
Q8. What is the difference between Sₙ and aₙ?
aₙ is the nth term (a single term). Sₙ is the sum of the first n terms. They are related: aₙ = Sₙ − Sₙ₋₁ for n ≥ 2.
