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If (S_n+S_{n-1}=1260) and (S_n-S_{n-1}=36), which is the value of (S_n)?

Answer and explanation

Correct answer: (648)

Direct answer: option B, \(S_n=648\). We are given two equations: \(S_n+S_{n-1}=1260\) and \(S_n-S_{n-1}=36\). Add them term by term: the two \(S_{n-1}\) terms cancel, giving \(2S_n=1296\). Divide by 2: \(S_n=648\). Option A, 630, is half of 1260 but ignores the second equation, so it is not justified. Option B is exactly the calculated value. Option C, 666, and option D, 684, do not satisfy the pair of equations; for example, each would make \(2S_n\) equal to 1332 or 1368 rather than 1296. Subtraction could also be used, but addition is quickest because it removes \(S_{n-1}\). Memory cue: when one equation has a plus and the other has a minus, adding them isolates the first quantity.

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SequencesProgressionsNatural-NumbersEquation

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