PYQs
Practise questions grouped by chapter and source year.
Practice questions are organised using the years and labels in our source collection. Work through the solutions with your teacher; these are not a complete official examination paper.
Practice question 1
Source year: 20253 marksFind sum of first 24 terms of AP: 5,8,11,14... [CBSE 2025]
Practice question 2
Source year: 20242 marksWhich term of AP 3,8,13,18... is 78? [CBSE 2024]
Practice question 3
Source year: 20253 marksFind sum of first 24 terms of AP: 5, 8, 11, 14, ... [CBSE 2025]
Practice question 4
Source year: 20242 marksWhich term of AP 3, 8, 13, 18, ... is 78? [CBSE 2024]
Practice question 5
Source year: 20233 marksIf 7th and 13th terms of AP are 34 and 64 respectively, find 18th term. [CBSE 2023]
Practice question 6
Source year: 20223 marksHow many three-digit numbers are divisible by 7? [CBSE 2022]
Practice question 7
Source year: 20212 marksFind d if a = 3 and a₁₀ = 30 for an AP. [CBSE 2021]
Formulae
Revisit important relationships and quick revision formulae.
nth term
aₙ = a + (n-1)d
Sum formula
Sₙ = n/2[2a + (n-1)d]
General Term (nth term)
aₙ = a + (n-1)d\nwhere a = first term, d = common difference
Sum of n terms
Sₙ = n/2 [2a + (n-1)d]\nor Sₙ = n/2 [a + l]\nwhere l = last term
Common Difference
d = a₂ - a₁ = aₙ - aₙ₋₁
Middle Term
If n is odd, middle term = (n+1)/2 th term\nIf n is even, middle terms = (n/2)th and (n/2 + 1)th terms
Derivations
Work through the reasoning behind key concepts.
Sum Formula Derivation
Let Sₙ = a + (a+d) + (a+2d) + ... + l. Write in reverse: Sₙ = l + (l-d) + (l-2d) + ... + a. Add both: 2Sₙ = (a+l) + (a+l) + ... + (a+l) = n(a+l). Therefore Sₙ = n/2(a+l) = n/2[2a + (n-1)d].
Sum Formula Derivation
Sₙ = a + (a+d) + (a+2d) + ... + l\nReverse: Sₙ = l + (l-d) + (l-2d) + ... + a\nAdding: 2Sₙ = n(a+l)\nTherefore: Sₙ = n/2 (a+l)
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