1. Chapter at a Glance
Reena's salary starts at ₹8,000 a month with an annual increment of ₹500. The steps of a staircase rise by the same height. Seats in an auditorium increase by 2 in each row. In all these, each term is obtained by adding a fixed number to the previous one. Such lists are arithmetic progressions (APs). In this chapter you will find any term of an AP and the sum of its first n terms using formulas — an important chapter that regularly carries 2-, 3- and 5-mark board questions.
- Definition of an AP; first term a and common difference d.
- nth term: aₙ = a + (n − 1)d.
- nth term from the end.
- Sum of first n terms: Sₙ = n/2[2a + (n − 1)d] = n/2(a + l).
- Relation aₙ = Sₙ − Sₙ₋₁.
- Applications: savings, salaries, seating, logs, ladders.
2. What Is an AP?
A list of numbers a₁, a₂, a₃, … is an arithmetic progression if the difference between any term and the previous term is the same: a₂ − a₁ = a₃ − a₂ = … = d, the common difference.
| List | Differences | AP? |
|---|---|---|
| 4, 10, 16, 22, … | 6, 6, 6 | Yes, d = 6 |
| 3, 1, −1, −3, … | −2, −2, −2 | Yes, d = −2 |
| 1, 1, 1, 1, … | 0, 0, 0 | Yes, d = 0 |
| 1, 4, 9, 16, … | 3, 5, 7 | No |
| √3, √12, √27, √48 | √3 each (= √3, 2√3, 3√3, 4√3) | Yes, d = √3 |
An AP with a last term is finite; otherwise it is infinite.
3. The nth Term
aₙ = a + (n − 1)d (aₙ is also written l when it is the last term)
Find the 10th term of 2, 7, 12, …
Solution: a = 2, d = 5. a₁₀ = 2 + 9 × 5 = 47.
Which term of 21, 18, 15, … is −81? Is 0 a term?
Solution: 21 + (n − 1)(−3) = −81 → n − 1 = 34 → n = 35. For 0: 21 − 3(n − 1) = 0 → n = 8. So 0 is the 8th term.
How many three-digit numbers are divisible by 7?
Solution: 105, 112, …, 994. 994 = 105 + (n − 1)7 → n = 128.
nth term from the end: for an AP with last term l, the nth term from the end = l − (n − 1)d.
4. Sum of the First n Terms
The young Gauss is said to have added 1 to 100 by writing the sum forwards and backwards: each pair adds to 101, and there are 100 pairs, so twice the sum is 100 × 101. The same trick works for any AP.
Sₙ = n/2 [2a + (n − 1)d]
Sₙ = n/2 (a + l), where l is the last (nth) term.
Also: aₙ = Sₙ − Sₙ₋₁.
Find the sum of the first 22 terms of 8, 3, −2, …
Solution: a = 8, d = −5. S₂₂ = 11[16 + 21(−5)] = 11(−89) = −979.
Show that 1 + 2 + … + n = n(n + 1)/2, and find 1 + 2 + … + 1000.
Solution: a = 1, l = n → Sₙ = n(1 + n)/2. For n = 1000: 500500.
If Sₙ = 4n − n², find the first term, the common difference and the nth term.
Solution: S₁ = 3 = a₁. S₂ = 4 → a₂ = 1. d = −2. aₙ = 3 + (n − 1)(−2) = 5 − 2n.
5. Applications
Shakila saves ₹5 on day 1, ₹10 on day 2, ₹15 on day 3, … How much in 30 days?
Solution: S₃₀ = 15[10 + 29 × 5] = 15 × 155 = ₹2,325.
200 logs are stacked with 20 in the bottom row, 19 in the next, 18 in the next, and so on. In how many rows are they placed, and how many logs are in the top row?
Solution: n/2[40 − (n − 1)] = 200 → n(41 − n) = 400 → n² − 41n + 400 = 0 → (n − 16)(n − 25) = 0. n = 25 gives a negative top row (20 − 24 = −4), so n = 16 rows; top row = 20 − 15 = 5 logs.
6. Solved Problems (Board Pattern)
The 3rd term of an AP is 16 and the 7th term exceeds the 5th by 12. Find the AP.
Solution: 2d = 12 → d = 6; a + 2d = 16 → a = 4. AP: 4, 10, 16, 22, …
Find the middle term of the AP 213, 205, 197, …, 37.
Solution: 37 = 213 + (n − 1)(−8) → n = 23. Middle term = 12th = 213 − 88 = 125.
How many terms of 9, 17, 25, … must be taken to give a sum of 636?
Solution: n/2[18 + 8(n − 1)] = 636 → 4n² + 5n − 636 = 0 → (n − 12)(4n + 53) = 0 → n = 12.
7. Master Formula Table
| Result | Formula |
|---|---|
| Common difference | d = aₙ₊₁ − aₙ |
| nth term | aₙ = a + (n − 1)d |
| nth term from end | l − (n − 1)d |
| Sum (first term and d) | Sₙ = n/2 [2a + (n − 1)d] |
| Sum (first and last) | Sₙ = n/2 (a + l) |
| Term from sums | aₙ = Sₙ − Sₙ₋₁ |
| Three terms in AP | a − d, a, a + d |
| a, b, c in AP | 2b = a + c |
8. Master Summary Table
| Question type | Method | Marks (typical) |
|---|---|---|
| Is it an AP? Find d | Check differences | 1 |
| Find nth term / which term | aₙ = a + (n − 1)d | 2 |
| Count terms (e.g. multiples) | Solve for n | 2–3 |
| Sum of n terms | Sₙ formulas | 3 |
| Real-life (savings, logs) | Model + Sₙ | 5 |
Chart — S₁₀ for three APs with a = 1:
9. Mind Map
- a, d
- 2b = a + c
- a + (n − 1)d
- From end: l − (n − 1)d
- n/2[2a + (n − 1)d]
- n/2(a + l)
- Savings, salary
- Logs, seats
10. Tips and Tricks
- If three numbers in AP have a given sum, take them as a − d, a, a + d — the sum is simply 3a.
- Use Sₙ = n/2(a + l) whenever the last term is known — less arithmetic.
- If a quadratic in n gives two values, check which makes sense (no negative row of logs).
- Using n instead of (n − 1) in aₙ.
- Computing d as a₁ − a₂.
- Counting terms: between 105 and 994 in steps of 7 there are 128 terms, not 127.
11. Important One-liners
- In an AP, the difference between consecutive terms is constant.
- d can be positive, negative or zero.
- a, b, c are in AP if and only if 2b = a + c.
- The nth term of an AP is a linear expression in n.
- Sₙ of an AP is a quadratic expression in n (when d ≠ 0).
- Sum of the first n odd numbers is n².
12. Exam-pattern Questions with Detailed Answers
The common difference of the AP 1/p, (1 − p)/p, (1 − 2p)/p, … is:
- p
- −p
- −1
- 1
Show answer
(1 − p)/p − 1/p = −p/p = −1.
Answer: (C)
Find the 11th term of −3, −1/2, 2, …
Show answer
d = 5/2. a₁₁ = −3 + 10 × 5/2 = 22.
Answer: 22
For what k are 2k + 1, 13, 5k − 3 in AP?
Show answer
26 = 2k + 1 + 5k − 3 → 7k = 28 → k = 4.
Answer: k = 4
Find the 20th term from the last of 3, 8, 13, …, 253.
Show answer
253 − 19 × 5 = 158.
Answer: 158
Find the sum of all odd numbers between 0 and 50.
Show answer
1, 3, …, 49: 25 terms. Sum = 25² = 625.
Answer: 625
The sum of the first n terms of an AP is 3n² + 5n. Find its 25th term.
Show answer
a₂₅ = S₂₅ − S₂₄ = (1875 + 125) − (1728 + 120) = 2000 − 1848 = 152.
Answer: 152
Which term of 3, 15, 27, 39, … will be 132 more than its 54th term?
Show answer
d = 12; 132 = 11 × 12, so it is 11 terms later: the 65th term.
Answer: 65th term
A sum of ₹700 is to be used for seven cash prizes, each ₹20 less than the previous one. Find each prize.
Show answer
7/2[2a − 120] = 700 → 2a − 120 = 200 → a = 160. Prizes: 160, 140, 120, 100, 80, 60, 40.
Answer: ₹160, 140, 120, 100, 80, 60, 40
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms.
Show answer
2a + 10d = 24, 2a + 14d = 44 → 4d = 20 → d = 5, a = −13. Terms: −13, −8, −3.
Answer: −13, −8, −3
A spiral is made of successive semicircles of radii 0.5, 1, 1.5, 2, … cm. What is the total length of 13 such semicircles? (π = 22/7)
Show answer
Lengths πr: π(0.5 + 1 + … + 6.5) = π × 13/2 × (0.5 + 6.5) = π × 45.5 = (22/7) × 45.5 = 143 cm.
Answer: 143 cm
The houses in a row are numbered 1 to 49. Show that there is a house x such that the sum of the numbers before it equals the sum after it, and find x.
Show answer
(x − 1)x/2 = [49 × 50/2] − x(x + 1)/2 → x² − x = 2450 − x² − x → 2x² = 2450 → x = 35.
Answer: x = 35
If the pth term of an AP is q and the qth term is p (p ≠ q), find its (p + q)th term.
Show answer
a + (p − 1)d = q, a + (q − 1)d = p. Subtract: (p − q)d = q − p → d = −1. a = q + p − 1. a_{p+q} = (p + q − 1) − (p + q − 1) = 0.
Answer: 0
13. Quick Revision Summary
- AP: constant difference d; a, b, c in AP ⇔ 2b = a + c.
- aₙ = a + (n − 1)d; from end: l − (n − 1)d.
- Sₙ = n/2[2a + (n − 1)d] = n/2(a + l).
- aₙ = Sₙ − Sₙ₋₁.
- Model savings, prizes, logs and seating; reject impossible values of n.