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If the sum of all consecutive integers from A to B (A<B), in

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If the sum of all consecutive integers from A to B (A<B), in  [#permalink]

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New post 24 Mar 2014, 04:04
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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)
E) 2S/(A+B)
C) (B−A)/2
D) S/(B−A+1)
E) (B−A)/(2S)

OE:
Sum of consecutive integers = Average x (largest - smallest + 1) -> S = M x (B - A + 1) -> S / (B - A + 1) = M
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Re: If the sum of all consecutive integers from A to B (A<B), in  [#permalink]

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New post 24 Mar 2014, 04:24
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\(Mean = \frac{sum of all the terms}{number of terms}\)
We already know the sum which is given as S.
We need the number of terms which needs to be calculated.

Since this is evenly placed set (set of consecutive numbers forms an AP with common difference of 1)

Number of terms in an AP is given by : \(\frac{Last term - first term}{common difference} + 1\)

Using this , we get the number of terms as : B - A + 1 (d = 1 in this case)

Hence the answer is: \(M = \frac{S}{(B - A + 1)}\)
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Re: If the sum of all consecutive integers from A to B (A<B), in  [#permalink]

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New post 03 Apr 2017, 06:16
Mean= Sum of the terms/Number of terms

B is the last term and A is the first term of the series.
Total number of terms= B-A+1
Sum= S

Mean (M)= S/(B-A+1)
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Re: If the sum of all consecutive integers from A to B (A<B), in &nbs [#permalink] 03 Apr 2017, 06:16
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