Class 10 Mathematics: Arithmetic Progressions important questions

Arithmetic Progressions is a chapter of Class 10 Mathematics for the Andhra Pradesh State Board. These 15 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 · 8 marks

   1. Check whether −150 is a term of the A.P. 11, 8, 5, 2, …

   a = 11, d = −3, so aₙ = 11 + (n − 1)(−3) = 14 − 3n. 14 − 3n = −150 gives 3n = 164 and n = 164/3, which is not a whole number. So −150 is not a term of the A.P.

2. Practice · 1 marks

   2. The nth term of an A.P. is aₙ = 2n − 6. Match: i) a₂ ii) a₃ iii) a₄ with p) 0 q) 2 r) −2 A) i–p, ii–r, iii–q B) i–r, ii–q, iii–p C) i–r, ii–p, iii–q D) i–q, ii–r, iii–p

   C) a₂ = 4 − 6 = −2 (r), a₃ = 6 − 6 = 0 (p), a₄ = 8 − 6 = 2 (q).

3. Important · 4 marks

   3. Find the sum of the odd numbers between 0 and 50.

   1, 3, 5, …, 49 is an A.P. with a = 1, d = 2. Number of terms: 49 = 1 + (n − 1) × 2, so n = 25. Sum = (25/2)(1 + 49) = 25 × 25 = 625.

4. Very Important · 8 marks

   4. The sum of the first 7 terms of an A.P. is 49 and the sum of the first 17 terms is 289. Find the sum of the first n terms.

   S₇ = (7/2)(2a + 6d) = 49, so a + 3d = 7 … (1) S₁₇ = (17/2)(2a + 16d) = 289, so a + 8d = 17 … (2) Subtracting: 5d = 10, so d = 2 and a = 1. Sₙ = (n/2)[2(1) + (n − 1)(2)] = (n/2)(2n) = n².

5. Practice · 1 marks

   5. Assertion (A): 4 is the common difference of the A.P. 5, 9, 13, …, 185. Reason (R): The common difference of an A.P. is d = aₙ − aₙ₋₁. A) Both A and R are true and R is the correct explanation of A. B) Both A and R are true but R is not the correct explanation of A. C) A is true, R is false. D) A is false, R is true.

   A) Both A and R are true, and R explains A: d = 9 − 5 = 13 − 9 = 4.

6. Important · 4 marks

   6. For an A.P. a₁, a₂, a₃, …, write the formula for each and name the terms in it: A) the nth term aₙ B) the sum of the first n terms Sₙ C) the common difference d

   A) aₙ = a + (n − 1)d, where a is the first term, d the common difference and n the term number. B) Sₙ = (n/2)[2a + (n − 1)d], or Sₙ = (n/2)(a + l) where l is the last (nth) term. C) d = a₂ − a₁ = aₙ − aₙ₋₁ (any term minus the term before it).

7. Very Important · 8 marks

   7. Students plan to plant trees to reduce air pollution. Each section of each class plants as many trees as its class number: a section of Class I plants 1 tree, Class II plants 2 trees, and so on up to Class XII. There are three sections in each class. How many trees will be planted?

   Trees per class (3 sections): 3, 6, 9, …, 36, an A.P. with a = 3, d = 3, n = 12. S₁₂ = (12/2)(3 + 36) = 6 × 39 = 234. 234 trees will be planted.

8. Practice · 1 marks

   8. Statement I: 3, 6, 9, 12, … is an A.P. Statement II: 2, 2, 2, 2, … is not an A.P. A) Both statements are true. B) Statement I is true, Statement II is false. C) Statement I is false, Statement II is true. D) Both statements are false.

   B) Statement I is true (d = 3). Statement II is false: 2, 2, 2, … is an A.P. with common difference 0.

9. Important · 4 marks

   9. State the formulae for the nth term and the sum of the first n terms of an A.P. Using them, find the 10th term and the sum of the first 10 terms of the A.P. 2, 7, 12, …

   aₙ = a + (n − 1)d and Sₙ = (n/2)[2a + (n − 1)d]. Here a = 2, d = 5. a₁₀ = 2 + 9 × 5 = 47. S₁₀ = (10/2)(2 + 47) = 5 × 49 = 245.

10. Practice · 1 marks

    10. Which of the following could represent an A.P.? A) 2, 4, 8, 6 B) 5, 5, 5, 5 C) 1, 4, 9, 16 D) 2, 3, 5, 8

    B) 5, 5, 5, 5 The difference between consecutive terms is always 0, so it is an A.P. with common difference d = 0. The others do not have a constant difference.

11. Important · 4 marks

    11. A spiral is made of successive semicircles, with centres alternately at A and B, starting with centre A, of radii 0.5 cm, 1 cm, 1.5 cm, 2 cm, and so on. What is the total length of a spiral made of 13 consecutive semicircles? (Take π = 22/7.)

    Length of a semicircle of radius r = πr. The radii form an A.P. 0.5, 1, 1.5, … with a = 0.5, d = 0.5, n = 13. Sum of radii = (13/2)[2(0.5) + 12(0.5)] = (13/2)(7) = 45.5 cm. Total length = π × 45.5 = (22/7) × 45.5 = 143 cm.

12. Very Important · 8 marks

    12. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

    d = a₃ − a₂ = 18 − 14 = 4, and a = a₂ − d = 14 − 4 = 10. S₅₁ = (51/2)[2a + 50d] = (51/2)[20 + 200] = (51/2)(220) = 51 × 110 = 5610.

13. Practice · 1 marks

    13. Your friend writes the sequence 5, 8, 11, 15, 17 as an A.P. What would you tell your friend?

    It is not an A.P. The differences are 3, 3, 4, 2, which are not constant. In an A.P. every term differs from the previous one by the same amount.

14. Important · 4 marks

    14. How many two-digit numbers are divisible by 3?

    The two-digit multiples of 3 are 12, 15, 18, …, 99: an A.P. with a = 12, d = 3, last term 99. 99 = 12 + (n − 1) × 3, so n − 1 = 29 and n = 30. There are 30 such numbers.

15. Very Important · 8 marks

    15. The sum of the first n terms of an A.P. is 4n − n². Find the first term, the sum of the first two terms, the second term, the third term, the 10th term and the nth term.

    Sₙ = 4n − n². S₁ = 4 − 1 = 3, so the first term a₁ = 3. S₂ = 8 − 4 = 4, so a₂ = S₂ − S₁ = 1. S₃ = 12 − 9 = 3, so a₃ = S₃ − S₂ = −1. Common difference d = 1 − 3 = −2. a₁₀ = a + 9d = 3 − 18 = −15. aₙ = 3 + (n − 1)(−2) = 5 − 2n.