If S_n = 4n² + 3n, find the sum from the 61st term to the 90th term.
Answer and explanation
Correct answer: 18090
For a sequence whose sum of the first n terms is S_n, the sum from the rth term through the sth term is S_s − S_{r−1}. Here the required range is from the 61st to the 90th term, so calculate S_90 − S_60. S_90 = 4(90²) + 3(90) = 4(8100) + 270 = 32670, while S_60 = 4(60²) + 3(60) = 4(3600) + 180 = 14580. Hence the required sum is 32670 − 14580 = 18090. Therefore, option A is correct. Subtracting S_61 or S_90 − S_61 would exclude or mishandle a boundary term, producing distractor values.
Frequently asked questions
What is the correct answer to this question?
18090
Why is this the correct answer?
For a sequence whose sum of the first n terms is S_n, the sum from the rth term through the sth term is S_s − S_{r−1}. Here the required range is from the 61st to the 90th term, so calculate S_90 − S_60. S_90 = 4(90²) + 3(90) = 4(8100) + 270 = 32670, while S_60 = 4(60²) + 3(60) = 4(3600) + 180 = 14580. Hence the required sum is 32670 − 14580 = 18090. Therefore, option A is correct. Subtracting S_61 or S_90 − S_61 would exclude or mishandle a boundary term, producing distractor values.
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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