CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 5 Arithmetic Progressions
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Boards begin February 2027
Pre-boards are in December
Arithmetic Progressions is in Algebra, which carries 20 of the 80 marks in the 2026-27 CBSE Maths paper
41 NCERT Exemplar questions from Chapter 5
Exercise 5.1
Exemplar Ex 5.1 Q11 mark
In an AP, if d = −4, n = 7 and aₙ = 4, then a is
Step 1aₙ = a + (n − 1)d ⇒ 4 = a + 6 × (−4) ⇒ 4 = a − 24.
Step-by-step solution
aₙ = a + (n − 1)d ⇒ 4 = a + 6 × (−4) ⇒ 4 = a − 24.
Justify whether it is true to say that −1, −3/2, −2, 5/2, ... forms an AP as a₂ − a₁ = a₃ − a₂.
Step 1[1 mark] a₂ − a₁ = −3/2 − (−1) = −1/2 and a₃ − a₂ = −2 − (−3/2) = −1/2, so the first two differences are equal — but equality of only these does not make a list an AP.
Step-by-step solution
[1 mark] a₂ − a₁ = −3/2 − (−1) = −1/2 and a₃ − a₂ = −2 − (−3/2) = −1/2, so the first two differences are equal — but equality of only these does not make a list an AP.
[1 mark] a₄ − a₃ = 5/2 − (−2) = 9/2 ≠ −1/2, so the differences are not all equal; hence the statement is false.
Answer:False — a₄ − a₃ = 9/2 ≠ a₃ − a₂, so it is not an AP.
Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?
Step 1[1 mark] The nth terms are (2 + (n − 1)d) and (7 + (n − 1)d); their difference is (7 + (n − 1)d) − (2 + (n − 1)d) = 7 − 2 = 5, independent of n.
Step-by-step solution
[1 mark] The nth terms are (2 + (n − 1)d) and (7 + (n − 1)d); their difference is (7 + (n − 1)d) − (2 + (n − 1)d) = 7 − 2 = 5, independent of n.
[1 mark] So every pair of corresponding terms differs by 5 — exactly the difference of the first terms — which is why the 10th, 21st and all corresponding differences are equal.
Answer:Because the difference of corresponding terms equals the difference of first terms (7 − 2 = 5) for every n.
The first term of an AP is −5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Step 1[1 mark] Sₙ = (n/2)(a + l) ⇒ 120 = (n/2)(−5 + 45) = 20n, so n = 6.
Step-by-step solution
[1 mark] Sₙ = (n/2)(a + l) ⇒ 120 = (n/2)(−5 + 45) = 20n, so n = 6.
[1 mark] l = a + (n − 1)d ⇒ 45 = −5 + 5d ⇒ 5d = 50.
Jaspal Singh repays a total loan of ₹1,18,000 by paying every month, starting with the first instalment of ₹1000, increasing by ₹100 every month. What amount will he pay in the 30th instalment? After paying the 30th instalment, how much loan does he still have to pay?
Step 1[1 mark] The monthly instalments form an AP with first term a = 1000 and common difference d = 100.
Step-by-step solution
[1 mark] The monthly instalments form an AP with first term a = 1000 and common difference d = 100.
The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
Step 1[1 mark] With 37 terms the middle term is the 19th; the three middle terms are a₁₈, a₁₉, a₂₀ with sum 3a₁₉ = 225, so a₁₉ = 75, i.e. a + 18d = 75.
Step-by-step solution
[1 mark] With 37 terms the middle term is the 19th; the three middle terms are a₁₈, a₁₉, a₂₀ with sum 3a₁₉ = 225, so a₁₉ = 75, i.e. a + 18d = 75.
[1 mark] The last three terms a₃₅, a₃₆, a₃₇ have sum 3a₃₆ = 429, so a₃₆ = 143, i.e. a + 35d = 143.
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and last to the product of the two middle terms is 7 : 15. Find the numbers.
Step 1Let the four numbers be a − 3d, a − d, a + d, a + 3d.
Step-by-step solution
Let the four numbers be a − 3d, a − d, a + d, a + 3d.
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Assertion-Reason1 mark
Assertion (A): The numbers 2, 5, 10, 17, … form an AP.
Reason (R): If aₙ is a quadratic polynomial in n, then the list a₁, a₂, a₃, … is not an AP because consecutive differences are not constant.
Step 1Differences: 5 − 2 = 3, 10 − 5 = 5, 17 − 10 = 7. Not constant, so the list is NOT an AP. Hence A is false.
Step-by-step solution
Differences: 5 − 2 = 3, 10 − 5 = 5, 17 − 10 = 7. Not constant, so the list is NOT an AP. Hence A is false.
aₙ = n² + 1 here, a quadratic in n; aₙ − aₙ₋₁ = 2n − 1 depends on n, so the list cannot be an AP — R is true.
Yasmeen saves ₹32 in the first month, ₹36 in the second month, ₹40 in the third month, and so on, each month saving ₹4 more than the previous month.
(i) Write the first term and common difference of this AP.
(ii) Write Yasmeen's savings in the 12th month.
(iii) In how many months will her total savings be ₹2000?
(iv) What will be her saving in the 20th month?
Step 1(i) a = 32, d = 4.
Step-by-step solution
(i) a = 32, d = 4.
(ii) a₁₂ = 32 + 11 × 4 = 32 + 44 = ₹76.
(iii) Sₙ = n/2 [2 × 32 + (n − 1) × 4] = n/2 × (60 + 4n) = n(30 + 2n) = 2000. So 2n² + 30n − 2000 = 0 ⇒ n² + 15n − 1000 = 0. Using the quadratic formula n = [−15 ± √(225 + 4000)] / 2 = [−15 ± 65]/2. Taking n > 0: n = 25 months.
(iv) a₂₀ = 32 + 19 × 4 = 32 + 76 = ₹108.
Answer:(i) a = 32, d = 4; (ii) ₹76; (iii) 25 months; (iv) ₹108.
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Questions from the NCERT Class 10 Mathematics Exemplar Problems (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.