CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 20

Exercise 3: Sum of First n Terms (Questions 3 to 10)

Question 1. In an AP:
(i) Given a = 5, d = 3, aₙ = 50, find n and Sₙ.
(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(v) Given d = 5, S₉ = 75, find a and a₉.
(vi) Given a = 2, d = 8, Sₙ = 90, find n and aₙ.
(vii) Given a = 8, aₙ = 62, Sₙ = 210, find n and d.
(viii) Given aₙ = 4, d = 2, Sₙ = −14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are 9 terms, find a.
Answer:
(i) 50 = 5 + (n − 1) × 3 ⇒ n = 16. S₁₆ = 8(5 + 50) = 440.
(ii) 35 = 7 + 12d ⇒ d = \( \frac{7}{3} \). S₁₃ = \( \frac{13}{2}(7+35) \) = 273.
(iii) 37 = a + 11 × 3 ⇒ a = 4. S₁₂ = 6(4 + 37) = 246.
(iv) a + 2d = 15 ...(i) and S₁₀ = 5(2a + 9d) = 125 ⇒ 2a + 9d = 25 ...(ii). From (i), 2a + 4d = 30. Subtracting from (ii): 5d = −5 ⇒ d = −1, a = 17, a₁₀ = 17 + 9(−1) = 8.
(v) S₉ = \( \frac{9}{2}(2a+8\times5) \) = 75 ⇒ 2a + 40 = \( \frac{50}{3} \) ⇒ a = \( -\frac{35}{3} \). a₉ = \( -\frac{35}{3}+8\times5=\frac{85}{3} \).
(vi) 90 = \( \frac{n}{2} \)[4 + (n − 1) × 8] ⇒ 4n² − 2n − 90 = 0 ⇒ (n − 5)(4n + 18) = 0. Since n is a natural number, n = 5 and a₅ = 2 + 4 × 8 = 34.
(vii) 210 = \( \frac{n}{2} \)(8 + 62) ⇒ n = 6. 62 = 8 + 5d ⇒ d = \( \frac{54}{5} \).
(viii) 4 = a + 2(n − 1) ⇒ a = 6 − 2n. Sₙ = \( \frac{n}{2} \)(a + 4) = −14 ⇒ n(10 − 2n) = −28 ⇒ n² − 5n − 14 = 0 ⇒ (n − 7)(n + 2) = 0. So n = 7 and a = −8.
(ix) 192 = 4[6 + 7d] ⇒ 6 + 7d = 48 ⇒ d = 6.
(x) 144 = \( \frac{9}{2} \)(a + 28) ⇒ a + 28 = 32 ⇒ a = 4.
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 2. How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?
Answer:
a = 9, d = 8. 636 = \( \frac{n}{2} \)[18 + 8(n − 1)] = n(4n + 5) ⇒ 4n² + 5n − 636 = 0 ⇒ (n − 12)(4n + 53) = 0.
n cannot be negative or fractional, so n = 12.
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 3. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer:
400 = \( \frac{n}{2} \)(5 + 45) ⇒ n = 16.
45 = 5 + 15d ⇒ d = \( \frac{8}{3} \). So there are 16 terms.
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 4. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer:
350 = 17 + (n − 1) × 9 ⇒ n − 1 = 37 ⇒ n = 38.
S = \( \frac{38}{2} \)(17 + 350) = 19 × 367 = 6973.
There are 38 terms and the sum is 6973.
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 5. Find the sum of the first 22 terms of an AP in which d = 7 and the 22nd term is 149.
Answer:
149 = a + 21 × 7 ⇒ a = 2.
S₂₂ = 11(2 + 149) = 1661.
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. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
d = 18 − 14 = 4 and a = 14 − 4 = 10.
S₅₁ = \( \frac{51}{2} \)[20 + 50 × 4] = \( \frac{51}{2}\times220 \) = 5610.
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. If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of the first n terms.
Answer:
S₇ = \( \frac{7}{2}(2a+6d) \) = 49 ⇒ a + 3d = 7 ...(i)
S₁₇ = \( \frac{17}{2}(2a+16d) \) = 289 ⇒ a + 8d = 17 ...(ii)
(ii) − (i): 5d = 10 ⇒ d = 2, a = 1.
Sₙ = \( \frac{n}{2}[2+(n-1)2]=n^2 \). So Sₙ = n².
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. Show that a₁, a₂, ..., aₙ, ... form an AP where aₙ is defined as below. Also find the sum of the first 15 terms in each case.
(i) aₙ = 3 + 4n   (ii) aₙ = 9 − 5n
Answer:
(i) a₁ = 7, a₂ = 11, a₃ = 15, a₄ = 19, ... and aₖ₊₁ − aₖ = 4 for every k, so it is an AP with a = 7, d = 4. S₁₅ = \( \frac{15}{2}[14+56] \) = 525.
(ii) a₁ = 4, a₂ = −1, a₃ = −6, a₄ = −11, ... and aₖ₊₁ − aₖ = −5 for every k, so it is an AP with a = 4, d = −5. S₁₅ = \( \frac{15}{2}[8-70] \) = −465.
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.