If the sum of all integers from $$A$$ to $$B$$ inclusive, where $$A \lt B$$, is $$S$$, and the average (arithmetic mean) of these integers is $$M$$, then which of the following is equal to $$M$$?

Incorrect.
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Incorrect.
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Incorrect.
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This is a trap answer choice and a good one. The average of a set of
consecutive integers is the average of its first and last terms, or $$\frac{B \color{red}{+} A}{2}$$ (rather than $$\frac{B \color{red}{-} A}{2}$$).
For example, if $$A=1$$ and $$B=3$$, the average of the set will be 2—while this answer choice is $$\frac{3-1}{2} = \frac{2}{2} = 1$$.
This is yet another example of why __Plugging In__ is the safer approach for variables questions—even if you're good with algebra. Under pressure and time limits, even the most proficient algebraist makes these careless mistakes, and once you move on to the next question, there's no turning back.

Correct.
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In this case, the sum is $$S$$ and the average is $$M$$. The number of terms is the difference between the extremes plus 1:
>$$\text{Number of terms} = B - A + 1$$
Thus, you get
>$$\text{Sum} = \text{Average} \times \text{Number of terms}$$
> $$S = M \times (B-A+1)$$
You need to solve for $$M$$. Divide both sides by $$B-A+1$$:
> $$\frac{S}{B-A+1} = M$$
**Alternative method:**
__Plug In__ easy-to-use numbers for $$A$$ and $$B$$ such as $$A=1$$ and $$B=3$$. The sum $$S= 1+2+3=6$$, and the average $$M=2$$. Your **goal value** is $$M=2$$. All answer choices other than D are eliminated.

Incorrect.
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$$\frac{2S}{A+B+1}$$

$$\frac{2S}{A+B}$$

$$\frac{B-A}{2}$$

$$\frac{S}{B-A+1}$$

$$\frac{B-A}{2S}$$