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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

Answer: D, 6765. S_n denotes the sum of the first n terms. To add terms from the rth through the sth term, subtract the sum before the rth term: S_s − S_{r−1}. Therefore, the required sum is S_45 − S_30, because the 31st term must be included and the first 30 terms must be removed. Calculate S_45 = 6(45²) + 45 = 6(2025) + 45 = 12,150 + 45 = 12,195. Next, S_30 = 6(30²) + 30 = 6(900) + 30 = 5,400 + 30 = 5,430. Thus the required sum is 12,195 − 5,430 = 6,765. Option A, B, and C do not equal this difference. A common boundary mistake is to use S_45 − S_31; that removes the 31st term as well, giving the wrong range. Memory cue: for terms r to s, always write S_s − S_{r−1}.

Related tags

Arithmetic ProgressionPartial SumsRange SumSequence CalculationFinding The Sum Of The First $N$ Terms Of An ApFinding The Sum Of The First N Terms Of An ApArithmetic Progressions (Ap)Arithmetic Progressions Ap

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