CBSE Class 10 · Maths · board exam 2027
Chapter 5: Arithmetic Progressions — notes
The chapter’s 7 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.
1. What is an AP?
This example.
- • Reena's salary (₹): 8000, 8500, 9000, 9500, …
- • 8500 − 8000 = 500
- • 9000 − 8500 = 500
- • 9500 − 9000 = 500
- • AP: every term = the term before it + a fixed number
- • Each number in the list = a term
- • first term = a, common difference = d
- • d = 2nd term − 1st term
- • Reena: a = 8000, d = 8500 − 8000 = 500
- • 1, 2, 3, 4, … d = 2 − 1 = 1 (positive)
- • 100, 70, 40, 10, … d = 70 − 100 = −30 (negative)
- • 3, 3, 3, 3, … d = 3 − 3 = 0 (zero)
- • Finite AP: has a last term, e.g. 2, 4, 6, 8, 10
- • Infinite AP: never ends, e.g. 5, 10, 15, 20, …
2. General form of an AP, and how to test a list
The first term is a.
- • General form: a, a + d, a + 2d, a + 3d, …
- • 1st term = a, 2nd term = a + d, 3rd term = a + 2d
- • a = 6, d = 3:
- • 6 + 3 = 9, 9 + 3 = 12, 12 + 3 = 15
- • AP: 6, 9, 12, 15, …
- • a = 4, d = −3:
- • 4 + (−3) = 1, 1 + (−3) = −2, −2 + (−3) = −5
- • AP: 4, 1, −2, −5, …
- • Test: (next term) − (term before it) must be the same every time
- • aₖ₊₁ − aₖ = d for every k
- • 4, 10, 16, 22, …
- • 10 − 4 = 6, 16 − 10 = 6, 22 − 16 = 6
- • AP with a = 4, d = 6
- • Next terms: 22 + 6 = 28, 28 + 6 = 34
- • 1, 1, 2, 3, 5: 1 − 1 = 0, 2 − 1 = 1 (not equal: not an AP)
- • 1, 4, 9, 16: 4 − 1 = 3, 9 − 4 = 5, 16 − 9 = 7 (not an AP)
- • √3, √12, √27, …
- • √12 = √(4 × 3) = 2√3
- • √27 = √(9 × 3) = 3√3
- • 2√3 − √3 = √3, 3√3 − 2√3 = √3
- • AP with d = √3
- • a, b, c in AP ⇒ b − a = c − b ⇒ 2b = a + c
- • 2k + 1, 13, 5k − 3 are in AP
- • 13 − (2k + 1) = (5k − 3) − 13
- • 12 − 2k = 5k − 16
- • 12 + 16 = 5k + 2k
- • 28 = 7k ⇒ k = 4
- • Terms: 9, 13, 17 (d = 4 ✓)
3. The nth term of an AP
We could write 15 numbers, but that is slow.
- • 1st year: 8000
- • 2nd year: 8000 + 1 × 500 = 8500
- • 3rd year: 8000 + 2 × 500 = 9000
- • 15th year: 8000 + 14 × 500 = 8000 + 7000 = 15000
- • Salary in the 15th year = ₹15000
- • a₁ = a
- • a₂ = a + d = a + (2 − 1)d
- • a₃ = a + 2d = a + (3 − 1)d
- • a₄ = a + 3d = a + (4 − 1)d
- • aₙ = a + (n − 1)d
- • aₙ = nth term (general term), last term l = aₙ
- • 2, 7, 12, …: a = 2, d = 7 − 2 = 5, n = 10
- • a₁₀ = 2 + (10 − 1) × 5
- • = 2 + 9 × 5 = 2 + 45 = 47
- • 10th term = 47
- • 10, 7, 4, …: a = 10, d = 7 − 10 = −3
- • a₃₀ = 10 + (30 − 1) × (−3)
- • = 10 + 29 × (−3) = 10 − 87 = −77
- • 30th term = −77
- • nth term from the end = l − (n − 1)d
- • Example: 3, 8, 13, …, 253. 20th term from the end?
- • l = 253, d = 8 − 3 = 5
- • 253 − (20 − 1) × 5 = 253 − 19 × 5 = 253 − 95 = 158
- • a₃ = a + 2d = 5 ...(1)
- • a₇ = a + 6d = 9 ...(2)
- • (2) − (1): 4d = 4 ⇒ d = 1
- • a + 2 × 1 = 5 ⇒ a = 3
- • AP: 3, 4, 5, 6, …
4. Which term is it? The formula used backwards
Sometimes the value of a term is given, and we must find its position.
- • 21, 18, 15, …: a = 21, d = 18 − 21 = −3
- • Let aₙ = −81
- • −81 = 21 + (n − 1)(−3)
- • −81 = 21 − 3n + 3
- • −81 = 24 − 3n
- • −81 − 24 = −3n
- • −105 = −3n
- • n = (−105)/(−3) = 35
- • −81 is the 35th term
- • 21 + (n − 1)(−3) = 0
- • 3(n − 1) = 21 ⇒ n − 1 = 7 ⇒ n = 8
- • Check: a₈ = 21 + 7 × (−3) = 21 − 21 = 0 ✓
- • 5, 11, 17, 23, …: a = 5, d = 6
- • 5 + (n − 1) × 6 = 301
- • 6n − 1 = 301 ⇒ 6n = 302
- • n = 302/6 = 501/3 (not a whole number)
- • 301 is not a term of this AP
- • 12, 15, 18, …, 99: a = 12, d = 3, aₙ = 99
- • 12 + (n − 1) × 3 = 99
- • (n − 1) × 3 = 87 ⇒ n − 1 = 29 ⇒ n = 30
- • There are 30 two-digit numbers divisible by 3
5. Sum of the first n terms
She puts ₹100 in her daughter's money box on her first birthday, ₹150 on the second, ₹200 on the third, and so on.
- • Shakila (₹): 100, 150, 200, 250, … total of 21 terms = ?
- • Gauss: S = 1 + 2 + 3 + … + 99 + 100
- • S = 1 + 2 + 3 + … + 100
- • S = 100 + 99 + 98 + … + 1
- • 2S = 101 + 101 + … + 101 (100 times)
- • 2S = 100 × 101 = 10100
- • S = 10100/2 = 5050
- • Sₙ = a + (a + d) + … + [a + (n − 1)d]
- • Sₙ = [a + (n − 1)d] + … + (a + d) + a
- • 2Sₙ = n × [2a + (n − 1)d]
- • Sₙ = n/2[2a + (n − 1)d]
- • 2a + (n − 1)d = a + [a + (n − 1)d] = a + l
- • Sₙ = n/2(a + l), l = last term
- • a = 100, d = 50, n = 21
- • S₂₁ = 21/2 × (2 × 100 + 20 × 50)
- • = 21/2 × (200 + 1000) = 21/2 × 1200
- • = 21 × 600 = 12600
- • Total = ₹12600
- • 8, 3, −2, …: a = 8, d = 3 − 8 = −5, n = 22
- • S₂₂ = 22/2 × (2 × 8 + 21 × (−5))
- • = 11 × (16 − 105) = 11 × (−89) = −979
- • 34 + 32 + 30 + … + 10: a = 34, d = −2, l = 10
- • 10 = 34 + (n − 1)(−2) ⇒ (n − 1) × 2 = 24 ⇒ n = 13
- • S₁₃ = 13/2 × (34 + 10) = 13/2 × 44 = 13 × 22 = 286
- • 1 + 2 + 3 + … + n = (n(n + 1))/2
- • 1 + 2 + … + 10 = (10 × 11)/2 = 55
6. When Sₙ is given: finding a, d and aₙ
Sometimes the question gives a formula for S n, and asks for the AP. Two ideas are enough.
- • a = S₁
- • aₙ = Sₙ − Sₙ₋₁
- • Sₙ = 4n − n²
- • S₁ = 4 × 1 − 1² = 3 ⇒ a = 3
- • S₂ = 4 × 2 − 2² = 8 − 4 = 4
- • a₂ = S₂ − S₁ = 4 − 3 = 1
- • d = a₂ − a₁ = 1 − 3 = −2
- • Sₙ₋₁ = 4(n − 1) − (n − 1)²
- • = 4n − 4 − (n² − 2n + 1)
- • = 4n − 4 − n² + 2n − 1 = 6n − n² − 5
- • aₙ = (4n − n²) − (6n − n² − 5)
- • = 4n − n² − 6n + n² + 5 = 5 − 2n
- • Check: a₁ = 5 − 2 = 3 ✓
- • a₁₀ = 5 − 2 × 10 = −15
- • aₙ = 3n − 2: a₁ = 1, a₂ = 4, a₃ = 7
- • aₙ₊₁ − aₙ = [3(n + 1) − 2] − [3n − 2] = 3
- • aₙ = pn + q ⇒ d = p
7. Finding n from the sum: a quadratic, and the double answer
A question where the sum is given and n is unknown.
- • 24, 21, 18, …: a = 24, d = −3, Sₙ = 78
- • 78 = n/2[2 × 24 + (n − 1)(−3)]
- • 78 = n/2(48 − 3n + 3)
- • 78 = n/2(51 − 3n)
- • 156 = 51n − 3n²
- • 3n² − 51n + 156 = 0
- • n² − 17n + 52 = 0
- • n² − 4n − 13n + 52 = 0
- • n(n − 4) − 13(n − 4) = 0
- • (n − 4)(n − 13) = 0
- • n − 4 = 0 ⇒ n = 4; n − 13 = 0 ⇒ n = 13
- • S₄ = 24 + 21 + 18 + 15 = 78
- • a₅ to a₁₃: 12 + 9 + 6 + 3 + 0 − 3 − 6 − 9 − 12 = 0
- • So S₁₃ = S₄ = 78: both n = 4 and n = 13
- • n must be a positive whole number
- • Negative or fraction value of n ⇒ rejected (write the reason)
Summary
Very good! An AP is a list where each term is the one before it plus a fixed number d. The first term is a, and the general form is a, a plus d, a plus 2d. To test a list, check that the next term minus the term before it stays the same. Then the nth term, a n equals a plus n minus 1 into d, which also tells us which term a number is. Then the sum, S n equals n by 2 into 2a plus n minus 1 into d, or n by 2 into a plus l. And when S n is given, the first term is S1 and a n is S n minus S n minus 1. Write these formulas once in your notebook without looking. Your homework is on your screen. In the next class we will practise board questions on this chapter, step by step. If anything was not clear, press the Ask button and ask me. Done, you worked very well today!
AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 5. Written from the textbook, checked twice.
Questions parents and students ask
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