Sum of AP Word Problems
The formula for the sum of first n terms of an AP has wide applications in real-life problems involving savings, distances, production, wages, and patterns.
These word problems require translating a real-world situation into AP language — identifying the first term (a), common difference (d), and number of terms (n) — then applying the sum formula.
Two sum formulas are used depending on available information:
- S_n = (n/2)[2a + (n − 1)d] — when a, d, and n are known.
- S_n = (n/2)(a + l) — when first term (a), last term (l), and n are known.
The sum formula transforms complex counting problems into simple algebraic calculations. Instead of adding 100 individual terms, the formula gives the answer in one step. Gauss famously used this idea as a child to instantly add the numbers from 1 to 100.
In CBSE board exams, AP sum word problems are among the most frequently tested questions, typically carrying 3-5 marks. Mastering the formula and its application to various contexts is essential.
What is Sum of AP Word Problems?
Definition: The sum of first n terms of an AP with first term a and common difference d is denoted by S_n.
Formulas:
S_n = (n/2)[2a + (n − 1)d]
S_n = (n/2)(a + l)
Where:
- a = first term
- d = common difference
- n = number of terms
- l = last term = a + (n − 1)d
When to Use Which Formula:
- Know a, d, n → Use S_n = (n/2)[2a + (n−1)d].
- Know a, l, n → Use S_n = (n/2)(a + l) (simpler).
- Finding n from S_n → Set up equation, which often becomes a quadratic in n.
- Finding a specific term from the sum → Use a_n = S_n − S_{n−1}.
Sum of AP Word Problems Formula
Key Relationships:
- nth term: a_n = a + (n − 1)d
- Sum: S_n = (n/2)[2a + (n − 1)d]
- Sum (using last term): S_n = (n/2)(a + l)
- nth term from sum: a_n = S_n − S_{n−1} (for n ≥ 2)
Common word problem patterns:
| Situation | a | d |
|---|---|---|
| Saving Rs 10, Rs 15, Rs 20, ... | 10 | 5 |
| Production: 500, 525, 550, ... | 500 | 25 |
| Logs stacked: 20, 19, 18, ... | 20 | −1 |
| Distance: 10 km, 12 km, 14 km, ... | 10 | 2 |
Important Derived Results:
| Series | AP Parameters | Sum Formula |
|---|---|---|
| 1 + 2 + 3 + ... + n | a = 1, d = 1 | n(n+1)/2 |
| 2 + 4 + 6 + ... + 2n | a = 2, d = 2 | n(n+1) |
| 1 + 3 + 5 + ... + (2n−1) | a = 1, d = 2 | n² |
| 5 + 10 + 15 + ... + 5n | a = 5, d = 5 | 5n(n+1)/2 |
Derivation and Proof
Strategy for solving AP word problems:
- Read carefully — identify what varies in a regular pattern.
- Find a: The first value in the sequence (first month's saving, first row's count, etc.).
- Find d: The constant increase (or decrease) from one term to the next.
- Find n: The number of terms (months, rows, days, etc.).
- Decide what to find: Is it a specific term (use a_n formula) or a total (use S_n formula)?
- Apply the formula and solve.
- Verify — check if the answer makes sense in context.
Gauss's Method (Historical Context):
- Young Gauss was asked to add 1 + 2 + 3 + ... + 100.
- He paired the numbers: (1 + 100), (2 + 99), (3 + 98), ..., (50 + 51).
- Each pair sums to 101. There are 50 pairs.
- Total = 50 × 101 = 5050.
- This is exactly S_n = (n/2)(a + l) with n = 100, a = 1, l = 100.
The general sum formula S_n = (n/2)(a + l) works on the same principle: pair the first and last terms, the second and second-last, etc. Each pair sums to (a + l), and there are n/2 such pairs.
Types and Properties
Common categories of AP sum word problems:
Type 1: Savings / Deposits
- A person saves a fixed amount more each month.
- Find total savings after n months.
Type 2: Production / Manufacturing
- A factory increases production by a fixed amount each week/month.
- Find total production over n periods.
Type 3: Stacking / Arrangement
- Objects stacked in rows with each row having one more (or less) than the previous.
- Find total objects in the stack.
Type 4: Distance / Motion
- Distance covered increases by a fixed amount each interval.
- Find total distance over n intervals.
Type 5: Reverse problems
- Given S_n, find n or d or a.
Recognising AP in Word Problems:
- Look for phrases like "increases by a fixed amount", "decreases by a constant", "each successive", "every month/day/week".
- If the increase is fixed (additive), it is an AP. If the increase is multiplicative (percentage), it is a GP (not AP).
- Common words indicating AP: "additional", "extra", "more than the previous", "less than the previous by a fixed amount".
Important Number Series as APs:
- Natural numbers: 1, 2, 3, ..., n → S = n(n+1)/2
- Even numbers: 2, 4, 6, ..., 2n → S = n(n+1)
- Odd numbers: 1, 3, 5, ..., (2n-1) → S = n²
- Multiples of k: k, 2k, 3k, ..., nk → S = kn(n+1)/2
Methods
Step-by-step method for AP sum word problems:
- Identify the AP: What quantity forms the AP? Write the first few terms.
- Extract a, d, n: From the problem statement.
- Choose the formula:
- If a, d, n are known → S_n = (n/2)[2a + (n − 1)d]
- If a, l, n are known → S_n = (n/2)(a + l)
- If finding n → set up an equation and solve
- Solve algebraically — some problems lead to quadratic equations.
- Check reasonableness: n must be a positive integer. Reject negative or fractional values of n.
Handling Quadratic Equations from AP Sum Problems:
- When S_n is given and n is unknown, the equation S_n = (n/2)[2a + (n−1)d] leads to a quadratic in n.
- Rearrange to standard form: an² + bn + c = 0.
- Solve using the quadratic formula or factorisation.
- Reject negative values of n (number of terms cannot be negative).
- Reject fractional values of n (you cannot have a fractional number of terms in a real-world context).
- If the problem says "how many terms are needed for the sum to exceed X", find the smallest integer n satisfying S_n > X.
Common Mistakes:
- Wrong sign on d: If the quantity decreases, d is negative. Forgetting the negative sign gives wrong results.
- Confusing n and a_n: n is the count of terms; a_n is the value of the nth term. They are different quantities.
- Off-by-one errors: The number of terms from the 5th to the 20th inclusive is 16 (not 15). Count carefully.
Solved Examples
Example 1: Monthly Savings Problem
Problem: Sachin saves Rs 100 in the first month and increases his savings by Rs 50 each month. Find his total savings in 2 years.
Solution:
Given:
- a = 100, d = 50, n = 24 (months in 2 years)
Using S_n = (n/2)[2a + (n − 1)d]:
- S₂₄ = (24/2)[2(100) + 23(50)]
- = 12[200 + 1150]
- = 12 × 1350
- = Rs 16,200
Example 2: Production Increase
Problem: A factory produces 500 widgets in the first week and increases production by 25 widgets every week. Find the total production in 20 weeks.
Solution:
Given:
- a = 500, d = 25, n = 20
S₂₀ = (20/2)[2(500) + 19(25)]:
- = 10[1000 + 475]
- = 10 × 1475
- = 14,750 widgets
Example 3: Stacking Logs
Problem: Logs are stacked with 20 in the bottom row, 19 in the next, 18 in the next, and so on. The top row has 1 log. Find the total number of logs.
Solution:
Given:
- a = 20, d = −1, last term l = 1
- n = ? Using a_n = a + (n−1)d: 1 = 20 + (n−1)(−1) → n = 20
S₂₀ = (20/2)(20 + 1):
- = 10 × 21
- = 210 logs
Example 4: Sum of Natural Numbers 1 to 100
Problem: Find the sum of first 100 natural numbers.
Solution:
Given:
- AP: 1, 2, 3, ..., 100. Here a = 1, l = 100, n = 100.
S₁₀₀ = (100/2)(1 + 100):
- = 50 × 101
- = 5050
Note: This is the famous result attributed to Gauss.
Example 5: Distance Covered by Runner
Problem: A runner covers 2 km on the first day and increases the distance by 500 m each day. How many days will it take to cover a total of 60 km?
Solution:
Given:
- a = 2 km, d = 0.5 km, S_n = 60 km
Using S_n = (n/2)[2a + (n − 1)d]:
- 60 = (n/2)[4 + 0.5(n − 1)]
- 60 = (n/2)[4 + 0.5n − 0.5]
- 60 = (n/2)(3.5 + 0.5n)
- 120 = n(3.5 + 0.5n)
- 120 = 3.5n + 0.5n²
- n² + 7n − 240 = 0
- (n + 19.25...)(n − 12.25...) — use quadratic formula
- n = (−7 + √(49 + 960))/2 = (−7 + √1009)/2 ≈ (−7 + 31.76)/2 ≈ 12.38
Since n must be a whole number, after 12 days the runner will have covered slightly less than 60 km. Check: S₁₂ = 6[4 + 5.5] = 6 × 9.5 = 57 km. After day 13: S₁₃ = 6.5[4 + 6] = 6.5 × 10 = 65 km.
Answer: The runner crosses 60 km during the 13th day.
Example 6: Penalty for Late Payment
Problem: A contractor is fined Rs 200 for the first day of delay, Rs 250 for the second, Rs 300 for the third, and so on. Find the total penalty for 30 days of delay.
Solution:
Given:
- a = 200, d = 50, n = 30
S₃₀ = (30/2)[2(200) + 29(50)]:
- = 15[400 + 1450]
- = 15 × 1850
- = Rs 27,750
Example 7: Rows of Seats in an Auditorium
Problem: An auditorium has 15 rows. The first row has 18 seats and each subsequent row has 2 more seats. Find the total seating capacity.
Solution:
Given:
- a = 18, d = 2, n = 15
Last row seats: l = 18 + 14 × 2 = 46
S₁₅ = (15/2)(18 + 46):
- = (15/2)(64)
- = 15 × 32
- = 480 seats
Example 8: Finding Number of Months to Save a Target
Problem: Reena saves Rs 32 in the first month and Rs 36 in the second month. If her savings increase in AP, how many months will it take for her total savings to reach Rs 2,000?
Solution:
Given:
- a = 32, d = 36 − 32 = 4, S_n = 2000
Setting up equation:
- 2000 = (n/2)[64 + (n − 1)4]
- 4000 = n[64 + 4n − 4]
- 4000 = n(60 + 4n)
- 4000 = 60n + 4n²
- n² + 15n − 1000 = 0
- n = (−15 + √(225 + 4000))/2 = (−15 + √4225)/2 = (−15 + 65)/2 = 25
Answer: Reena needs 25 months.
Example 9: Sum of Multiples
Problem: Find the sum of all multiples of 7 between 100 and 500.
Solution:
Finding the AP:
- First multiple of 7 ≥ 100: 105 (since 100/7 ≈ 14.28, so 15 × 7 = 105)
- Last multiple of 7 ≤ 500: 497 (since 500/7 ≈ 71.43, so 71 × 7 = 497)
- a = 105, l = 497, d = 7
Finding n:
- 497 = 105 + (n − 1)(7) → 392 = 7(n − 1) → n − 1 = 56 → n = 57
S₅₇ = (57/2)(105 + 497):
- = (57/2)(602)
- = 57 × 301
- = 17,157
Example 10: Decreasing AP — Salary Deductions
Problem: An employee's monthly bonus starts at Rs 5000 and decreases by Rs 200 each month. After how many months will the total bonus received be Rs 30,000?
Solution:
Given:
- a = 5000, d = −200, S_n = 30000
Equation:
- 30000 = (n/2)[10000 + (n − 1)(−200)]
- 60000 = n[10000 − 200n + 200]
- 60000 = n(10200 − 200n)
- 60000 = 10200n − 200n²
- 200n² − 10200n + 60000 = 0
- n² − 51n + 300 = 0
- n = (51 ± √(2601 − 1200))/2 = (51 ± √1401)/2 = (51 ± 37.43)/2
- n = 44.2 or n = 6.78
Since the bonus decreases, it reaches 0 at month 26 (5000 − 200 × 25 = 0). After that, no more bonus. So n = 6.78 means during the 7th month, the total crosses Rs 30,000.
Check: S₆ = 3[10000 + 5(−200)] = 3 × 9000 = 27000. S₇ = 3.5[10000 + 6(−200)] = 3.5 × 8800 = 30800.
Answer: During the 7th month, total bonus crosses Rs 30,000.
Real-World Applications
Banking and Finance:
- Recurring deposits with increasing amounts, EMI schedules, and compound saving plans often involve AP sums.
Construction:
- Calculating the number of bricks in a tapering wall, logs in a stack, or steps in a staircase.
Sports and Fitness:
- Training schedules where distance, repetitions, or weights increase by a fixed amount each day/week.
Business:
- Projecting total production when output increases steadily, or total revenue when sales grow linearly.
Education:
- A student reads 20 pages on the first day and 5 more pages each subsequent day. The total pages read after n days is an AP sum.
Nature:
- A plant grows 2 cm in the first week, 3 cm in the second, 4 cm in the third, etc. The total height after n weeks uses the AP sum formula.
Music:
- Some musical scales and rhythmic patterns follow arithmetic progressions in their frequency ratios or beat patterns.
Architecture:
- A pyramid of blocks has n blocks in the bottom row, (n-1) in the next, and so on. The total blocks = sum of first n natural numbers = n(n+1)/2.
Key Points to Remember
- S_n = (n/2)[2a + (n − 1)d] when a, d, n are known.
- S_n = (n/2)(a + l) when first and last terms are known.
- Always identify a, d, and n clearly from the word problem before applying the formula.
- If d is negative, the AP is decreasing — terms eventually reach zero or become negative.
- When finding n from S_n, you may get a quadratic equation. Reject negative or non-integer solutions.
- The nth term from sum: a_n = S_n − S_{n−1}.
- For sums of specific patterns: sum of first n natural numbers = n(n + 1)/2.
- Always verify by computing a few terms manually.
- Check that the answer is physically reasonable (positive number of items, sensible time period).
- AP sum word problems are high-frequency CBSE questions carrying 3–5 marks.
- The sum of first n odd numbers is always a perfect square: 1 + 3 + 5 + ... + (2n−1) = n².
- For sums of multiples: sum of first n multiples of k = k × n(n+1)/2.
Practice Problems
- A person saves Rs 200 in the first month, Rs 250 in the second, Rs 300 in the third, and so on. Find the total savings in 3 years.
- Find the sum of all multiples of 9 between 50 and 500.
- A ladder has rungs decreasing in length from bottom to top. Bottom rung = 50 cm, each rung 2 cm shorter. If there are 25 rungs, find the total length of wood needed.
- A factory makes 50 items on the first day and 5 more each succeeding day. Find the total after 20 days.
- 200 logs need to be stacked with 20 in the bottom row, 19 in the next, and so on. How many rows will be needed?
- An employee gets a starting salary of Rs 15,000 with Rs 500 annual increment. Find total salary earned in 20 years.
- A student reads 30 pages on day 1 and 5 more pages each subsequent day. In how many days will the total pages read reach 500?
- Find the sum of all three-digit numbers divisible by 11.
Frequently Asked Questions
Q1. When do I use S_n = (n/2)(a + l) vs S_n = (n/2)[2a + (n-1)d]?
Use S_n = (n/2)(a + l) when you know the first term (a), last term (l), and number of terms (n). Use S_n = (n/2)[2a + (n-1)d] when you know a, d, and n but not the last term.
Q2. What if the word problem leads to a quadratic equation for n?
Solve the quadratic using the quadratic formula or factorisation. Since n represents the number of terms (months, days, rows), reject any negative or fractional value. Only accept positive integer values.
Q3. How do I identify a and d from a word problem?
The first term (a) is the initial value (first month's saving, first day's production, bottom row count). The common difference (d) is the constant increase or decrease stated in the problem.
Q4. Can d be negative in word problems?
Yes. If the quantity decreases by a fixed amount each period (e.g., each row has one fewer log, salary decreases by a fixed penalty), then d is negative.
Q5. What is the sum of first n natural numbers?
S = n(n + 1)/2. This is the AP sum formula with a = 1 and d = 1. For example, 1 + 2 + 3 + ... + 100 = 100 × 101/2 = 5050.
Q6. How do I find the sum of all multiples of k between two numbers?
Find the first multiple of k ≥ lower limit (this is a). Find the last multiple of k ≤ upper limit (this is l). Common difference d = k. Find n using l = a + (n-1)d. Then apply S_n = (n/2)(a + l).
Q7. What if both a and d are unknown?
The problem will give two conditions (e.g., S_5 = 100 and a_10 = 50). Set up two equations using the AP formulas, solve simultaneously for a and d.
Q8. How is a_n related to S_n?
a_n = S_n − S_{n−1} for n ≥ 2. The nth term equals the difference between the sum of first n terms and the sum of first (n-1) terms. For n = 1, a_1 = S_1.
Q9. Can the sum of an AP be negative?
Yes, if the terms become negative and their negative sum exceeds the positive sum of earlier terms. For example, the AP 5, 2, −1, −4, ... has S_5 = 5 + 2 + (−1) + (−4) + (−7) = −5.
Q10. Is this topic important for board exams?
Extremely. AP word problems using the sum formula are among the most frequently asked 3-5 mark questions. Practice problems on savings, production, stacking, and distance are essential.
