Physics and Mathematics
Example 1 – Sum of A.P.
Practice Questions with Step-by-Step Solutions
Question 1. Find the sum of first 25 terms of the AP [3, 7, 11, …].
Step-by-Step Solution:
a = 3, d = 4, n = 25
[S₂₅ = \dfrac{25}{2} [6 + 96]]
[S₂₅ = \dfrac{25}{2} × 102 = 1275]
Conclusion:
The sum is 1275.
Question 2. Find the sum of first 15 terms of the AP [−2, −5, −8, …].
Step-by-Step Solution:
a = −2, d = −3, n = 15
[S₁₅ = \dfrac{15}{2} [−4 − 42]]
[S₁₅ = −345]
Conclusion:
The sum is −345.
Question 3. Find the sum of the first 25 terms of the AP: 3, 7, 11, 15, …
Step-by-Step Solution:
First term [a = 3]
Common difference [d = 7 − 3 = 4]
Number of terms [n = 25]
Use the formula
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[S₂₅ = \dfrac{25}{2} [2(3) + 24 × 4]]
Simplify:
[S₂₅ = \dfrac{25}{2} [6 + 96]][ = \dfrac{25}{2} × 102]
Final answer:
[S₂₅ = 1275]
Conclusion: The sum of the first 25 terms is [1275].
Question 4. Find the sum of the first 30 terms of the AP: 5, 3, 1, −1, …
Step-by-Step Solution:
First term [a = 5]
Common difference [d = 3 − 5 = −2]
Number of terms [n = 30]
Use the sum formula:
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[S₃₀ = \dfrac{30}{2} [10 + 29(−2)]]
Simplify:
[S₃₀ = 15 [10 − 58] = 15(−48)]
Final answer:
[S₃₀ = −720]
Conclusion: The sum of the first 30 terms is [−720].
Question 5. Find the sum of the first 40 natural numbers.
Step-by-Step Solution:
Natural numbers form an AP: 1, 2, 3, …
First term [a = 1]
Common difference [d = 1]
Number of terms [n = 40]
Apply formula:
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[S₄₀ = \dfrac{40}{2} [2 + 39]]
Simplify:
[S₄₀ = 20 × 41 = 820]
Conclusion: The sum of the first 40 natural numbers is [820].
Question 6. Find the sum of the first 15 terms of the AP whose first term is 8 and common difference is −3.
Step-by-Step Solution:
First term [a = 8]
Common difference [d = −3]
Number of terms [n = 15]
Use the sum formula:
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[S₁₅ = \dfrac{15}{2} [16 + 14(−3)]]
Simplify:
[S₁₅ = \dfrac{15}{2} [16 − 42]][ = \dfrac{15}{2} (−26)]
Final answer:
[S₁₅ = −195]
Conclusion: The sum of the first 15 terms is [−195].
Question 7. Find the sum of the first 20 even natural numbers.
Step-by-Step Solution:
Even numbers form an AP: 2, 4, 6, …
First term [a = 2]
Common difference [d = 2]
Number of terms [n = 20]
Apply formula:
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[S₂₀ = \dfrac{20}{2} [4 + 38]]
Simplify:
[S₂₀ = 10 × 42 = 420]
Conclusion: The sum of the first 20 even natural numbers is [420].
Question 8. Find the sum of the first 12 terms of the AP: −5, −2, 1, 4, …
Step-by-Step Solution:
First term [a = −5]
Common difference [d = 3]
Number of terms [n = 12]
Use sum formula:
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[S₁₂ = \dfrac{12}{2} [−10 + 11 × 3]]
Simplify:
[S₁₂ = 6 [−10 + 33]][ = 6 × 23]
Final answer:
[S₁₂ = 138]
Conclusion: The sum of the first 12 terms is [138].
Question 9. Find the sum of the first 50 terms of the AP: 7, 14, 21, …
Step-by-Step Solution:
First term [a = 7]
Common difference [d = 7]
Number of terms [n = 50]
Apply formula:
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[S₅₀ = \dfrac{50}{2} [14 + 49 × 7]]
Simplify:
[S₅₀ = 25 [14 + 343]][ = 25 × 357]
Final answer:
[S₅₀ = 8925]
Conclusion: The sum of the first 50 terms is [8925].
Question 10. How many terms of the AP: 6, 13, 20, … must be taken so that their sum is 406?
Step-by-Step Solution:
First term [a = 6]
Common difference [d = 7]
Let number of terms = [n]
Use sum formula:
[Sₙ = \dfrac{n}{2} [2a + (n − 1)d]]
Substitute values:
[406 = \dfrac{n}{2} [12 + 7n − 7]]
Simplify:
[406 = \dfrac{n}{2} (7n + 5)]
Multiply both sides by 2:
[812 = n(7n + 5)]
Solve:
[7n² + 5n − 812 = 0]
Solving gives [n = 11]
Conclusion: The required number of terms is [11].
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