CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 18

Exercise 1: Identifying and Writing APs

Question 1. In which of the following situations does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes \( \frac{1}{4} \) of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.
Answer:
(i) Fare for 1 km, 2 km, 3 km, 4 km, ... is Rs. 15, 23, 31, 39, ... Each term is 8 more than the previous one, so the list forms an AP with d = 8.
(ii) Let the air be V litres. After the 1st stroke it is \( \frac{3V}{4} \), after the 2nd \( \left(\frac{3}{4}\right)^2V \), after the 3rd \( \left(\frac{3}{4}\right)^3V \). The differences \( -\frac{V}{4} \) and \( -\frac{3V}{16} \) are not equal, so it is not an AP.
(iii) Cost for 1, 2, 3, 4 metres is Rs. 150, 200, 250, 300, ... The difference is always 50, so it forms an AP.
(iv) The amounts are \( 10000\left(1+\frac{8}{100}\right),\ 10000\left(1+\frac{8}{100}\right)^2,\ 10000\left(1+\frac{8}{100}\right)^3,\ \dots \), which have no common difference, so it is not an AP.
In simple words: A list is an AP only when the same amount is added every time; multiplying by a fixed number gives a different pattern.

Teacher's Note:
1. Add the same amount each step: AP. Multiply by the same factor each step: not an AP.
2. Compound interest always gives a list with a common ratio, not a common difference.

Question 2. Write the first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10   (ii) a = −2, d = 0   (iii) a = 4, d = −3   (iv) a = −1, d = \( \frac{1}{2} \)   (v) a = −1.25, d = −0.25
Answer:
(i) 10, 20, 30, 40.
(ii) −2, −2, −2, −2.
(iii) 4, 1, −2, −5.
(iv) \( -1,\ -\frac{1}{2},\ 0,\ \frac{1}{2} \).
(v) −1.25, −1.50, −1.75, −2.00.
In simple words: Start with a and keep adding d.

Teacher's Note:
1. The terms are a, a + d, a + 2d, a + 3d.
2. If d = 0 every term is the same.

Question 3. For the following APs, write the first term and the common difference:
(i) 3, 1, −1, −3, ...   (ii) −5, −1, 3, 7, ...   (iii) \( \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots \)   (iv) 0.6, 1.7, 2.8, 3.9, ...
Answer:
(i) a = 3, d = 1 − 3 = −2.
(ii) a = −5, d = −1 − (−5) = 4.
(iii) a = \( \frac{1}{3} \), d = \( \frac{5}{3}-\frac{1}{3}=\frac{4}{3} \).
(iv) a = 0.6, d = 1.7 − 0.6 = 1.1.
In simple words: The first term is the first number; d is second term minus first term.

Teacher's Note:
1. First find the first term a and the common difference d = a₂ − a₁, then use aₙ = a + (n − 1)d.
2. Check that the difference is the same between every pair of consecutive terms before calling a list an AP.

Question 4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
Answer:
(i) 2, 4, 8, 16: 4 − 2 = 2 but 8 − 4 = 4. Not an AP.
(ii) 2, \( \frac{5}{2} \), 3, \( \frac{7}{2} \): the difference is the same every time, d = \( \frac{1}{2} \). It is an AP. Next three terms: \( 4,\ \frac{9}{2},\ 5 \).
(iii) −1.2, −3.2, −5.2, −7.2: the difference is the same every time, d = −2. It is an AP. Next three terms: −9.2, −11.2, −13.2.
(iv) −10, −6, −2, 2: the difference is the same every time, d = 4. It is an AP. Next three terms: 6, 10, 14.
(v) 3, \( 3+\sqrt{2} \), \( 3+2\sqrt{2} \), \( 3+3\sqrt{2} \): the difference is the same every time, d = \( \sqrt{2} \). It is an AP. Next three terms: \( 3+4\sqrt{2},\ 3+5\sqrt{2},\ 3+6\sqrt{2} \).
(vi) 0.2, 0.22, 0.222, 0.2222: 0.22 − 0.2 = 0.02 but 0.222 − 0.22 = 0.002. Not an AP.
(vii) 0, −4, −8, −12: the difference is the same every time, d = −4. It is an AP. Next three terms: −16, −20, −24.
(viii) \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \): the difference is the same every time, d = 0. It is an AP. Next three terms: \( -\frac{1}{2},\ -\frac{1}{2},\ -\frac{1}{2} \).
(ix) 1, 3, 9, 27: 3 − 1 = 2 but 9 − 3 = 6. Not an AP.
(x) a, 2a, 3a, 4a: the difference is the same every time, d = a. It is an AP. Next three terms: 5a, 6a, 7a.
(xi) a, a², a³, a⁴: a² − a ≠ a³ − a² in general (they are equal only when a = 0 or 1). Not an AP.
(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \): the difference is the same every time, d = \( \sqrt{2} \). It is an AP. Next three terms: \( \sqrt{50},\ \sqrt{72},\ \sqrt{98} \).
(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12} \): \( \sqrt{6}-\sqrt{3}\neq\sqrt{9}-\sqrt{6} \). Not an AP.
(xiv) 1², 3², 5², 7²: 9 − 1 = 8 but 25 − 9 = 16. Not an AP.
(xv) 1², 5², 7², 73: the difference is the same every time, d = 24. It is an AP. Next three terms: 97, 121, 145.
In simple words: Find a₂ − a₁, a₃ − a₂ and a₄ − a₃; if all are equal the list is an AP.

Teacher's Note:
1. Always test all three differences, not just the first two.
2. For parts (x) and (xi) the letter a stands for a fixed non-zero number.

Exercise 2: nth Term of an AP

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 6. Choose the correct choice in the following and justify:
(i) 30th term of the AP 10, 7, 4, ... is (a) 97 (b) 77 (c) −77 (d) −87.
(ii) 11th term of the AP \( -3, -\frac{1}{2}, 2, \dots \) is (a) 28 (b) 22 (c) −38 (d) \( -48\frac{1}{2} \).
Answer:
(i) a = 10, d = 7 − 10 = −3. a₃₀ = 10 + 29 × (−3) = 10 − 87 = −77. Correct choice: (c).
(ii) a = −3, d = \( -\frac{1}{2}+3=\frac{5}{2} \). a₁₁ = \( -3+10\times\frac{5}{2}=22 \). Correct choice: (b).
In simple words: Substitute the given values in aₙ = a + (n − 1)d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 7. For the following APs, find the missing terms in the boxes:
(i) 2, __, 26   (ii) __, 13, __, 3   (iii) 5, __, __, \( 9\frac{1}{2} \)   (iv) −4, __, __, __, __, 6   (v) __, 38, __, __, __, −22
Answer:
(i) The middle term is the average: (2 + 26) ÷ 2 = 14.
(ii) a₂ = 13 and a₄ = 3 give 2d = 3 − 13 = −10, so d = −5. Then a₁ = 13 + 5 = 18 and a₃ = 13 − 5 = 8.
(iii) a = 5 and a₄ = \( \frac{19}{2} \) give 3d = \( \frac{19}{2}-5=\frac{9}{2} \), so d = \( \frac{3}{2} \). The missing terms are \( \frac{13}{2} \) and 8.
(iv) a = −4 and a₆ = 6 give 5d = 10, so d = 2. The missing terms are −2, 0, 2, 4.
(v) a₂ = 38 and a₆ = −22 give 4d = −60, so d = −15. The missing terms are a₁ = 53, a₃ = 23, a₄ = 8, a₅ = −7.
In simple words: Find d from the two known terms, then fill the gaps by adding or subtracting d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 8. Which term of the AP 3, 8, 13, 18, ... is 78?
Answer:
a = 3, d = 5. Let aₙ = 78: 3 + (n − 1) × 5 = 78 ⇒ n − 1 = 15 ⇒ n = 16.
Hence 78 is the 16th term.
In simple words: Substitute the given values in aₙ = a + (n − 1)d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 9. Find the number of terms in each of the following APs:
(i) 7, 13, 19, ..., 205   (ii) \( 18, 15\frac{1}{2}, 13, \dots, -47 \)
Answer:
(i) a = 7, d = 6. 205 = 7 + (n − 1) × 6 ⇒ n − 1 = 33 ⇒ n = 34.
(ii) a = 18, d = \( -\frac{5}{2} \). −47 = 18 + (n − 1)\( \left(-\frac{5}{2}\right) \) ⇒ n − 1 = 26 ⇒ n = 27.
In simple words: Substitute the given values in aₙ = a + (n − 1)d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 10. Check whether −150 is a term of the AP 11, 8, 5, 2, ...
Answer:
a = 11, d = −3. Suppose −150 = 11 + (n − 1)(−3) ⇒ −161 = −3(n − 1) ⇒ n − 1 = \( \frac{161}{3} \) ⇒ n = \( \frac{164}{3} \).
The number of terms must be a natural number, so −150 is not a term of this AP.
In simple words: Substitute the given values in aₙ = a + (n − 1)d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 11. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Answer:
a + 10d = 38 and a + 15d = 73. Subtracting: 5d = 35 ⇒ d = 7, and a = 38 − 70 = −32.
a₃₁ = −32 + 30 × 7 = 178.
In simple words: Substitute the given values in aₙ = a + (n − 1)d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 12. An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
Answer:
a + 2d = 12 and a + 49d = 106. Subtracting: 47d = 94 ⇒ d = 2, and a = 12 − 4 = 8.
a₂₉ = 8 + 28 × 2 = 64.
In simple words: Substitute the given values in aₙ = a + (n − 1)d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 13. If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero?
Answer:
a + 2d = 4 and a + 8d = −8. Subtracting: 6d = −12 ⇒ d = −2, and a = 8.
Let aₙ = 0: 8 + (n − 1)(−2) = 0 ⇒ n − 1 = 4 ⇒ n = 5. The 5th term is zero.
In simple words: Substitute the given values in aₙ = a + (n − 1)d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.

Question 14. If the 17th term of an AP exceeds its 10th term by 7, find the common difference.
Answer:
a₁₇ − a₁₀ = (a + 16d) − (a + 9d) = 7d = 7.
Hence the common difference is d = 1.
In simple words: The difference of two terms that are k places apart is k × d.

Teacher's Note:
1. Write what is given in terms of a, d, n, aₙ and Sₙ and pick the formula that has only the unknowns left.
2. When a quadratic in n appears, reject a negative or fractional value because the number of terms must be a natural number.