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If the sum of all consecutive integers from A to B (A<B), inclusive
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Bunuel
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If the sum of all consecutive integers from A to B (A<B), inclusive [#permalink] New post  26 May 2022, 05:45
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If the sum of all consecutive integers from A to B (A<B), inclusive, is S and the average (arithmetic mean) of these integers is M, then, in terms of A, B, and S, what is the value of M?

A. 2S/(A+B+1)
B. 2S/(A+B)
C. (B−A)/2
D. S/(B−A+1)
E. (B−A)/(2S)
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ThatDudeKnows
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Re: If the sum of all consecutive integers from A to B (A<B), inclusive [#permalink] New post  26 May 2022, 05:50
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Bunuel wrote:
If the sum of all consecutive integers from A to B (A<B), inclusive, is S and the average (arithmetic mean) of these integers is M, then, in terms of A, B, and S, what is the value of M?

A. 2S/(A+B+1)
B. 2S/(A+B)
C. (B−A)/2
D. S/(B−A+1)
E. (B−A)/(2S)



Any time there are variables in the question that are repeated in the answer choices, I look to see if there's an opportunity to Plug In.

Let's make A=1 and B=4. S=10 and M=2.5. Plug those into the answer choices.
(A) 20/6 Wrong
(B) 20/5 Wrong
(C) 3/2 Wrong
(D) 10/4 Keep it
(E) 3/20 Wrong

Answer choice D.


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If the sum of all consecutive integers from A to B (A<B), inclusive [#permalink] New post  26 May 2022, 06:13
Sum of a consecutive series = No. of terms * Average

We are given Average = \(M\), No. of terms \(= B - A + 1\), and Sum = \(S\).

Thus,

\(S = (B - A + 1)* M \)

\(M = \frac{S}{(B - A + 1)}\)
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If the sum of all consecutive integers from A to B (A<B), inclusive [#permalink]
26 May 2022, 06:13
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