CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 19

Exercise 2: nth Term of an AP (continued)

Question 1. Which term of the AP 3, 15, 27, 39, ... will be 132 more than its 54th term?
Answer:
a = 3, d = 12. a₅₄ = 3 + 53 × 12 = 639, so the required term is 639 + 132 = 771.
3 + (n − 1) × 12 = 771 ⇒ n − 1 = 64 ⇒ n = 65. It is the 65th term.
(Shortcut: 132 ÷ 12 = 11 places further than the 54th, i.e. 54 + 11 = 65.)
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 2. Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?
Answer:
Let the first terms be a and a′ and the common difference d. Then a₁₀₀ − a′₁₀₀ = (a + 99d) − (a′ + 99d) = a − a′ = 100.
a₁₀₀₀ − a′₁₀₀₀ = (a + 999d) − (a′ + 999d) = a − a′ = 100.
In simple words: Equal common differences mean the gap between the two APs never changes.

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 3. How many three-digit numbers are divisible by 7?
Answer:
The three-digit multiples of 7 are 105, 112, 119, ..., 994, an AP with a = 105 and d = 7.
994 = 105 + (n − 1) × 7 ⇒ n − 1 = 127 ⇒ n = 128.
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 4. How many multiples of 4 lie between 10 and 250?
Answer:
The multiples are 12, 16, 20, ..., 248, an AP with a = 12 and d = 4.
248 = 12 + (n − 1) × 4 ⇒ n − 1 = 59 ⇒ n = 60.
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 5. For what value of n are the nth terms of the two APs 63, 65, 67, ... and 3, 10, 17, ... equal?
Answer:
First AP: aₙ = 63 + (n − 1) × 2 = 61 + 2n. Second AP: aₙ = 3 + (n − 1) × 7 = 7n − 4.
61 + 2n = 7n − 4 ⇒ 5n = 65 ⇒ n = 13.
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 6. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer:
a₇ − a₅ = 2d = 12 ⇒ d = 6. a + 2d = 16 ⇒ a = 4.
The AP is 4, 10, 16, 22, ...
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. Find the 20th term from the last term of the AP 3, 8, 13, ..., 253.
Answer:
Reverse the AP: 253, 248, 243, ... with a = 253 and d = −5. The number of terms is (253 − 3) ÷ 5 + 1 = 51.
The 20th term from the last = 253 + 19 × (−5) = 253 − 95 = 158.
In simple words: The nth term from the end is l − (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 8. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Answer:
(a + 3d) + (a + 7d) = 2a + 10d = 24 ⇒ a + 5d = 12 ...(i)
(a + 5d) + (a + 9d) = 2a + 14d = 44 ⇒ a + 7d = 22 ...(ii)
(ii) − (i): 2d = 10 ⇒ d = 5, and a = 12 − 25 = −13.
The first three terms are −13, −8, −3.
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. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Answer:
The salaries 5000, 5200, 5400, ... form an AP with a = 5000, d = 200.
7000 = 5000 + (n − 1) × 200 ⇒ n − 1 = 10 ⇒ n = 11.
His salary reached Rs. 7000 in the 11th year, that is in 2005.
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. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week her weekly savings become Rs. 20.75, find n.
Answer:
a = 5, d = 1.75, aₙ = 20.75. 20.75 = 5 + (n − 1) × 1.75 ⇒ n − 1 = 15.75 ÷ 1.75 = 9.
Hence n = 10.
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.

Exercise 3: Sum of First n Terms (Questions 1 and 2)

Question 11. Find the sum of the following APs:
(i) 2, 7, 12, ..., to 10 terms   (ii) −37, −33, −29, ..., to 12 terms   (iii) 0.6, 1.7, 2.8, ..., to 100 terms   (iv) \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots \), to 11 terms
Answer:
(i) a = 2, d = 5, n = 10: S₁₀ = 5[4 + 45] = 245.
(ii) a = −37, d = 4, n = 12: S₁₂ = 6[−74 + 44] = −180.
(iii) a = 0.6, d = 1.1, n = 100: S₁₀₀ = 50[1.2 + 108.9] = 5505.
(iv) a = \( \frac{1}{15} \), d = \( \frac{1}{12}-\frac{1}{15}=\frac{1}{60} \), n = 11: S₁₁ = \( \frac{11}{2}\left[\frac{2}{15}+\frac{10}{60}\right]=\frac{11}{2}\times\frac{3}{10}=\frac{33}{20} \).
In simple words: Use Sₙ = n/2 [2a + (n − 1)d] or Sₙ = n/2 (a + l).

Teacher's Note:
1. Translate the story into a first term and a common difference before using any formula.
2. Use Sₙ = n/2 [2a + (n − 1)d] when d is known and Sₙ = n/2 (a + l) when the last term is known.

Question 12. Find the sums given below:
(i) \( 7 + 10\frac{1}{2} + 14 + \dots + 84 \)   (ii) 34 + 32 + 30 + ... + 10   (iii) −5 + (−8) + (−11) + ... + (−230)
Answer:
(i) a = 7, d = \( \frac{7}{2} \), l = 84. 84 = 7 + (n − 1)\( \frac{7}{2} \) ⇒ n = 23. S = \( \frac{23}{2}(7+84)=\frac{2093}{2}=1046\frac{1}{2} \).
(ii) a = 34, d = −2, l = 10. 10 = 34 + (n − 1)(−2) ⇒ n = 13. S = \( \frac{13}{2}(34+10)=286 \).
(iii) a = −5, d = −3, l = −230. −230 = −5 + (n − 1)(−3) ⇒ n = 76. S = 38(−5 − 230) = −8930.
In simple words: Use Sₙ = n/2 [2a + (n − 1)d] or Sₙ = n/2 (a + l).

Teacher's Note:
1. Translate the story into a first term and a common difference before using any formula.
2. Use Sₙ = n/2 [2a + (n − 1)d] when d is known and Sₙ = n/2 (a + l) when the last term is known.