Class 10 Mathematics: Arithmetic Progressions important questions

Arithmetic Progressions is a chapter of Class 10 Mathematics for the CBSE. These 12 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Very Important · 4 marks

   1. An auditorium for at least 1500 people has chairs in concentric circular rows, each row having 10 seats more than the previous one. (i) If the first row has 30 seats, how many seats are in the 10th row? (ii) For 1500 seats, how many rows are needed? OR If 1500 seats are to be arranged, how many seats are still left to be put after the 10th row? (iii) If there were 17 rows, how many seats would be in the middle row?

   (i) a₁₀ = 30 + 9 × 10 = 120. (ii) (n/2)[60 + 10(n − 1)] = 1500 gives n² + 5n − 300 = 0, (n + 20)(n − 15) = 0, so n = 15 rows. OR S₁₀ = 5(60 + 90) = 750, so 1500 − 750 = 750 seats remain. (iii) The middle row is the 9th: a₉ = 30 + 8 × 10 = 110.

2. Practice · 1 marks

   2. Two APs have the same common difference. The first term of one is −1 and of the other is −8. The difference between their 4th terms is: (a) 1 (b) −7 (c) 7 (d) 9

   (c) 7 With the same d, the difference between any corresponding terms equals the difference of the first terms: −1 − (−8) = 7.

3. Practice · 1 marks

   3. Assertion (A): −5, −5/2, 0, 5/2, … is in Arithmetic Progression. Reason (R): The terms of an Arithmetic Progression cannot have both positive and negative rational numbers. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.

   C. A is true (common difference 5/2) but R is false: an AP can contain both positive and negative terms.

4. Very Important · 4 marks

   4. Sanjitha's shot-put throw was initially 7.56 m, and she improved it by 9 cm every week. During a 15-day camp she started with 40 throws and increased the number of throws by 12 every day. (i) How many throws did she practise on the 11th day of the camp? (ii) What would her throw distance be at the end of 6 weeks? OR When will she achieve a throw of 11.16 m? (iii) How many throws did she do during the entire camp of 15 days?

   (i) t₁₁ = 40 + 10 × 12 = 160 throws. (ii) 7.56 + 6 × 0.09 = 8.10 m. OR 11.16 = 7.56 + 0.09k gives k = 40, so after 40 weeks (the 41st term of the A.P.). (iii) S₁₅ = (15/2)[80 + 14 × 12] = (15/2) × 248 = 1860 throws.

5. Practice · 1 marks

   5. If the nth term of an A.P. is 7n − 4, find its common difference.

   7 aₙ − aₙ₋₁ = (7n − 4) − (7(n − 1) − 4) = 7.

6. Very Important · 4 marks

   6. In a store, glass jars are arranged in layers: 3 jars in the top layer, 6 in the next, 9 in the third, and so on till the 8th layer. (i) Write the A.P. of jars in each layer from the top, and its common difference. (ii) Can a layer have 34 jars if this pattern continues? Justify. (iii) (A) Find an expression for the total number of jars in n layers, and hence S₈. OR (B) If the shopkeeper adds 3 jars to each layer, how many jars are in the 5th layer?

   (i) 3, 6, 9, 12, …, 24; d = 3. (ii) 34 = 3 + (n − 1)3 gives n = 34/3, not a whole number, so no. (iii) (A) Sₙ = (n/2)[6 + 3(n − 1)] = 3n(n + 1)/2, so S₈ = 3 × 8 × 9/2 = 108. OR (B) The A.P. becomes 6, 9, 12, …, so t₅ = 6 + 4 × 3 = 18.

7. Important · 2 marks

   7. The A.P. 8, 10, 12, … has 60 terms. Find the sum of its last 10 terms.

   a = 8, d = 2. The 51st term = 8 + 50 × 2 = 108 and the 60th term = 8 + 59 × 2 = 126. Sum of the last 10 terms = (10/2)(108 + 126) = 1170.

8. Important · 2 marks

   8. Find the middle term of the A.P. 6, 13, 20, …, 230.

   230 = 6 + (n − 1)7 gives n = 33, so the middle term is the 17th: 6 + 16 × 7 = 118.

9. Very Important · 4 marks

   9. Aryan writes the A.P. −5, −2, 1, 4, … and Roshan writes 187, 184, 181, … (i) Find the sum of the common differences of the two progressions. (ii) Find the 34th term of Roshan's progression. (iii) (A) Find the sum of the first 10 terms of Aryan's progression. OR (B) Which term of the two progressions has the same value?

   (i) 3 + (−3) = 0. (ii) a₃₄ = 187 + 33(−3) = 88. (iii) (A) S₁₀ = (10/2)[2(−5) + 9 × 3] = 5 × 17 = 85. OR (B) −5 + 3(n − 1) = 187 − 3(n − 1) gives 6(n − 1) = 192, so n = 33: the 33rd terms are equal.

10. Practice · 1 marks

    10. The middle term of the A.P. 10, 7, 4, …, −62 is: (A) −26 (B) −29 (C) −32 (D) −35

    (A) −26 −62 = 10 + (n − 1)(−3) gives n = 25, so the middle term is the 13th: a₁₃ = 10 + 12(−3) = −26.

11. Important · 3 marks

    11. Ridhi drew a polygon with n sides. The smallest exterior angle is 8°, and each next exterior angle is 4° more than the previous one. Find the number of sides.

    The exterior angles form an A.P. with a = 8 and d = 4, and their sum is 360°. (n/2)[16 + (n − 1)4] = 360, so n² + 3n − 180 = 0, i.e. (n + 15)(n − 12) = 0. n = 12 (n cannot be negative). The polygon has 12 sides.

12. Important · 3 marks

    12. Find the sum of the integers between 1 and 400 that are multiples of both 4 and 5.

    These are multiples of 20: 20, 40, …, 380 (a = 20, d = 20). 380 = 20 + (n − 1)20 gives n = 19. S₁₉ = (19/2)(20 + 380) = 3800.