If the sum of an AP is S_n = 6n² + n, find the sum from the 31st term to the 45th term.
Answer and explanation
Correct answer: 6765
When S_n is given, the sum from the rth term to the sth term is S_s − S_{r−1}. Therefore, the required sum is S₄₅ − S₃₀. Calculate S₄₅ = 6(45²) + 45 = 6(2025) + 45 = 12,195. Also, S₃₀ = 6(30²) + 30 = 6(900) + 30 = 5,430. Hence the range sum is 12,195 − 5,430 = 6,765. Thus option D is correct. Subtracting S₃₁ would omit the 31st term, while using S₄₅ alone would include all earlier terms as well.
Frequently asked questions
What is the correct answer to this question?
6765
Why is this the correct answer?
When S_n is given, the sum from the rth term to the sth term is S_s − S_{r−1}. Therefore, the required sum is S₄₅ − S₃₀. Calculate S₄₅ = 6(45²) + 45 = 6(2025) + 45 = 12,195. Also, S₃₀ = 6(30²) + 30 = 6(900) + 30 = 5,430. Hence the range sum is 12,195 − 5,430 = 6,765. Thus option D is correct. Subtracting S₃₁ would omit the 31st term, while using S₄₅ alone would include all earlier terms as well.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
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