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In an AP, a = 7, and the sum from the 21st term to the 40th term is 3680. Find d.

Answer and explanation

Correct answer: 6

Answer: A, d = 6. The terms from the 21st through the 40th positions contain 40 − 21 + 1 = 20 terms. The 21st term is a + 20d = 7 + 20d, and the 40th term is a + 39d = 7 + 39d. The sum of these 20 AP terms is number of terms divided by 2 multiplied by the sum of the first and last terms: 3680 = 20/2[(7 + 20d) + (7 + 39d)] = 10(14 + 59d). Divide by 10: 368 = 14 + 59d. Hence 59d = 354 and d = 6. Therefore option A is correct. Option B, C, and D fail when substituted into the same equation. The common confusion is using 20d for the last term; 20 is the number of selected terms, whereas the 40th term has 39 differences after the first term. A useful check is that S_40 − S_20 gives the same block sum.

Related tags

Arithmetic ProgressionCommon DifferenceRange SumAp FormulaFinding 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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