CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 21

Exercise 3: Sum of First n Terms (Questions 11 to 20)

Question 1. If the sum of the first n terms of an AP is 4n − n², what is the first term (that is S₁)? What is the sum of the first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.
Answer:
a₁ = S₁ = 4(1) − 1 = 3. S₂ = 4(2) − 4 = 4, so a₂ = S₂ − S₁ = 1.
d = 1 − 3 = −2, so aₙ = 3 + (n − 1)(−2) = 5 − 2n.
a₃ = 5 − 6 = −1 and a₁₀ = 5 − 20 = −15.
In simple words: The nth term is Sₙ − Sₙ₋₁.

Teacher's Note:
1. Use aₙ = Sₙ − Sₙ₋₁ for n ≥ 2 and a₁ = S₁.
2. Verify with the formula aₙ = 5 − 2n for n = 1.

Question 2. Find the sum of the first 40 positive integers divisible by 6.
Answer:
The integers are 6, 12, 18, ..., an AP with a = 6, d = 6.
S₄₀ = 20[12 + 39 × 6] = 20 × 246 = 4920.
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 3. Find the sum of the first 15 multiples of 8.
Answer:
a = 8, d = 8. S₁₅ = \( \frac{15}{2}[16+14\times8] \) = \( \frac{15\times128}{2} \) = 960.
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 4. Find the sum of the odd numbers between 0 and 50.
Answer:
The odd numbers 1, 3, 5, ..., 49 form an AP with a = 1, d = 2, l = 49.
49 = 1 + (n − 1) × 2 ⇒ n = 25. S = \( \frac{25}{2}(1+49) \) = 625.
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 5. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days?
Answer:
The penalties form an AP with a = 200 and d = 50.
S₃₀ = 15[400 + 29 × 50] = 15 × 1850 = 27750.
The contractor has to pay Rs. 27750.
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 6. A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer:
The prizes form an AP with first term a (the largest prize) and d = −20, and S₇ = 700.
700 = \( \frac{7}{2} \)[2a + 6(−20)] ⇒ 2a − 120 = 200 ⇒ a = 160.
The prizes are Rs. 160, 140, 120, 100, 80, 60 and 40.
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 7. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be the same as the class in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Answer:
The numbers 1, 2, 3, ..., 12 form an AP with a = 1, d = 1. Trees planted by one section of every class = S₁₂ = 6(2 + 11) = 78.
Three sections of each class: 3 × 78 = 234 trees.
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 8. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown in the figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
[Hint: The lengths of successive semicircles are l₁, l₂, l₃, l₄, ... with centres at A, B, A, B, ...]
Answer:
The length of a semicircle of radius r is πr. So l₁ = \( \frac{\pi}{2} \), l₂ = π, l₃ = \( \frac{3\pi}{2} \), l₄ = 2π, ... which is an AP with a = \( \frac{\pi}{2} \) and d = \( \frac{\pi}{2} \).
S₁₃ = \( \frac{13}{2}\left[\pi+12\times\frac{\pi}{2}\right]=\frac{13}{2}\times7\pi=\frac{13}{2}\times7\times\frac{22}{7} \) = 143 cm.
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 9. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Answer:
The rows have 20, 19, 18, ... logs: an AP with a = 20, d = −1. Let Sₙ = 200.
200 = \( \frac{n}{2} \)[40 + (n − 1)(−1)] ⇒ n² − 41n + 400 = 0 ⇒ (n − 16)(n − 25) = 0.
n = 25 would give a₂₅ = 20 − 24 = −4 logs, which is impossible, so n = 16. The top row has a₁₆ = 20 − 15 = 5 logs.
In simple words: Use Sₙ = n/2 [2a + (n − 1)d] or Sₙ = n/2 (a + l).

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. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3).]
Answer:
The potatoes are 5, 8, 11, 14, ... m from the bucket. For each potato she runs to it and back, so the distances are 2 × 5, 2 × 8, 2 × 11, ... = 10, 16, 22, 28, ..., an AP with a = 10, d = 6, n = 10.
S₁₀ = 5[20 + 9 × 6] = 5 × 74 = 370.
The competitor has to run 370 m.
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.

Exercise 4: Optional Questions

Question 11. Which term of the AP 121, 117, 113, ... is its first negative term? [Hint: Find n for aₙ < 0.]
Answer:
a = 121, d = −4, so aₙ = 121 − 4(n − 1) = 125 − 4n.
aₙ < 0 ⇒ 125 − 4n < 0 ⇒ n > 31.25. The smallest such n is 32.
The 32nd term is the first negative term (a₃₂ = −3).
In simple words: Use Sₙ = n/2 [2a + (n − 1)d] or Sₙ = n/2 (a + l).

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. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.
Answer:
a₃ + a₇ = 2a + 8d = 6 ⇒ a = 3 − 4d. Also a₃ = a + 2d = 3 − 2d and a₇ = a + 6d = 3 + 2d.
a₃ × a₇ = (3 − 2d)(3 + 2d) = 9 − 4d² = 8 ⇒ d² = \( \frac{1}{4} \) ⇒ d = ± \( \frac{1}{2} \).
If d = \( \frac{1}{2} \): a = 1 and S₁₆ = 8[2 + 15 × \( \frac{1}{2} \)] = 76.
If d = \( -\frac{1}{2} \): a = 5 and S₁₆ = 8[10 − 7.5] = 20.
In simple words: Use Sₙ = n/2 [2a + (n − 1)d] or Sₙ = n/2 (a + l).

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. A ladder has rungs 25 cm apart (see the figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \( 2\frac{1}{2} \) m apart, what is the length of the wood required for the rungs?
[Hint: Number of rungs = \( \frac{250}{25}+1 \).]
Answer:
Distance between the top and bottom rungs = 250 cm, so the number of rungs is \( \frac{250}{25}+1 \) = 11.
The lengths of the rungs decrease uniformly, so they form an AP with a = 45, l = 25, n = 11.
Length of wood = S₁₁ = \( \frac{11}{2}(45+25) \) = 385 cm.
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 14. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint: Sₓ₋₁ = S₄₉ − Sₓ]
Answer:
The sum of the numbers of the houses before house x is Sₓ₋₁ = \( \frac{(x-1)x}{2} \) and the sum after it is S₄₉ − Sₓ = \( 1225-\frac{x(x+1)}{2} \).
Equating: \( \frac{x(x-1)}{2}=1225-\frac{x(x+1)}{2} \) ⇒ x² = 1225 ⇒ x = ±35.
The house number cannot be negative, so x = 35 (such a house exists).
In simple words: Use Sₙ = n/2 [2a + (n − 1)d] or Sₙ = n/2 (a + l).

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 15. A small terrace at a football ground comprises 15 steps, each of which is 50 m long and built of solid concrete. Each step has a rise of \( \frac{1}{4} \) m and a tread of \( \frac{1}{2} \) m (see the figure). Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step = \( \frac{1}{4}\times\frac{1}{2}\times50 \) m³.]
Answer:
The heights of the steps are \( \frac{1}{4},\frac{1}{2},\frac{3}{4},\dots \) m, and each step has tread \( \frac{1}{2} \) m and length 50 m, so the volumes are \( \frac{25}{4},\frac{25}{2},\frac{75}{4},\dots \) m³, an AP with a = \( \frac{25}{4} \) and d = \( \frac{25}{4} \).
S₁₅ = \( \frac{15}{2}\left[\frac{25}{2}+14\times\frac{25}{4}\right]=\frac{15}{2}\times100=750 \).
The volume of concrete required is 750 m³.
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.