Physics and Mathematics
Example 3 Sum of A.P.
Practice Questions with Step-by-Step Solutions
Practice Question 1. If the sum of 5 terms of an A.P. is 25 and the sum of 15 terms is 225, find the sum of n terms.
Step-by-Step Solution:
Given: [S₅ = 25], [S₁₅ = 225]
Formula: [Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
For [n = 5]:
3. [\dfrac{5}{2} (2a + 4d) = 25]
4. [2a + 4d = 10] → Equation (1)
For [n = 15]:
5. [\dfrac{15}{2} (2a + 14d) = 225]
6. [2a + 14d = 30] → Equation (2)
Subtract (1) from (2):
[10d = 20 ⇒ d = 2]
Substitute in (1):
[2a + 8 = 10 ⇒ a = 1]
Sum of n terms:
[Sₙ = \dfrac{n}{2} [2(1) + (n − 1)2]]
Simplify:
[Sₙ = n²]
Conclusion: The sum of n terms is [n²].
Practice Question 2. If the sum of 4 terms of an AP is 40 and the sum of 10 terms is 190, find the sum of n terms.
Step-by-Step Solution:
[S₄ = 40], [S₁₀ = 190]
For [n = 4]:
2. [\dfrac{4}{2} (2a + 3d) = 40]
3. [2a + 3d = 20] → (1)
For [n = 10]:
4. [\dfrac{10}{2} (2a + 9d) = 190]
5. [2a + 9d = 38] → (2)
Subtract (1) from (2):
[6d = 18 ⇒ d = 3]
Substitute in (1):
[2a + 9 = 20 ⇒ a = \dfrac{11}{2}]
General sum:
[Sₙ = \dfrac{n}{2} [11 + 3(n − 1)]]
Conclusion: Required sum obtained.
Practice Question 3. If the sum of 6 terms of an AP is 48 and the sum of 14 terms is 224, find the sum of n terms.
Step-by-Step Solution:
[S₆ = 48], [S₁₄ = 224]
[\dfrac{6}{2} (2a + 5d) = 48 ⇒ 2a + 5d = 16] → (1)
[\dfrac{14}{2} (2a + 13d) = 224 ⇒ 2a + 13d = 32] → (2)
Subtract:
[8d = 16 ⇒ d = 2]
Substitute:
[2a + 10 = 16 ⇒ a = 3]
[Sₙ = \dfrac{n}{2} [6 + 2(n − 1)]]
Conclusion: Sum of n terms obtained.
Practice Question 4. If the sum of 7 terms of an AP is 91 and the sum of 13 terms is 273, find the sum of n terms.
Step-by-Step Solution:
[S₇ = 91], [S₁₃ = 273]
[2a + 6d = 26] → (1)
[2a + 12d = 42] → (2)
Subtract:
[6d = 16 ⇒ d = \dfrac{8}{3}]
Substitute to get [a]
Write [Sₙ] formula.
Conclusion: Required expression obtained.
Practice Question 5. If the sum of 9 terms of an AP is 90 and the sum of 21 terms is 462, find the sum of n terms.
Step-by-Step Solution:
[S₉ = 90], [S₂₁ = 462]
[2a + 8d = 20]
[2a + 20d = 44]
[12d = 24 ⇒ d = 2]
[a = 2]
[Sₙ = \dfrac{n}{2} [4 + 2(n − 1)]]
Conclusion: Sum of n terms obtained.
Practice Question 6. If the sum of 10 terms of an AP is 150 and the sum of 20 terms is 500, find the sum of n terms.
Step-by-Step Solution:
[2a + 9d = 30]
[2a + 19d = 50]
[10d = 20 ⇒ d = 2]
[a = 6]
[Sₙ = \dfrac{n}{2} [12 + 2(n − 1)]]
Conclusion: Required result obtained.
Practice Question 7. If the sum of 8 terms of an AP is 64 and the sum of 19 terms is 361, find the sum of n terms.
Step-by-Step Solution:
[2a + 7d = 16]
[2a + 18d = 38]
[11d = 22 ⇒ d = 2]
[a = 1]
[Sₙ = n²]
Conclusion: Same type as the given image question.
Practice Question 8. If the sum of 12 terms of an AP is 180 and the sum of 30 terms is 900, find the sum of n terms.
Step-by-Step Solution:
[2a + 11d = 30]
[2a + 29d = 60]
[18d = 30 ⇒ d = \dfrac{5}{3}]
Find [a]
Conclusion: Required sum obtained.
Practice Question 9. If the sum of 6 terms of an AP is 18 and the sum of 18 terms is 162, find the sum of n terms.
Step-by-Step Solution:
[2a + 5d = 6]
[2a + 17d = 18]
[12d = 12 ⇒ d = 1]
[a = \dfrac{1}{2}]
Write [Sₙ] formula.
Conclusion: Sum of n terms obtained.
Practice Question 10. If the sum of 11 terms of an AP is 132 and the sum of 25 terms is 650, find the sum of n terms.
Step-by-Step Solution:
[2a + 10d = 24]
[2a + 24d = 52]
[14d = 28 ⇒ d = 2]
[a = 2]
[Sₙ = \dfrac{n}{2} [4 + 2(n − 1)]]
Conclusion: Required sum obtained.
Sign in with Google