Equations: One-Variable Equations

If the average (arithmetic mean) of $$k-2$$, $$k-1$$, $$k$$, $$k+1$$ is $$k$$, what is the value of $$k$$?

Correct. [[snippet]] Sum = $$k-2 + k-1 + k + k+1 = 4k-2$$ Number of things is 4 (four terms). Thus, the average of the four terms is >$$\displaystyle \frac {4k-2}{4} = \frac{4k}{4} - \frac{2}{4} = k- \frac{1}{2}$$. Now, according to the question, the average of the four terms is $$k$$, so we get $$k- \frac{1}{2}=k$$, which is equivalent to $$-\frac{1}{2}=0$$. This isn't true, no matter what $$k$$ you choose. Therefore, there is no value of $$k$$ that fits the information.

Incorrect.

[[snippet]] Carefully check your work. By the average formula, the average of the four terms is >$$\mbox{Average} = \frac{(k-2) + (k-1) + (k) + (k+1)}{4} = \frac{4k-2}{4}$$

Incorrect.

[[snippet]] Carefully check your calculations.

Incorrect.

[[snippet]] Make sure you set the average of the four terms equal to $$k$$ and solve for $$k$$.

Incorrect.

[[snippet]] By the average formula, the average of the four terms is >$$\mbox{Average} = \frac{(k-2) + (k-1) + (k) + (k+1)}{4} = \frac{4k-2}{4}$$

$$-1$$

$$-\frac{1}{4}$$

$$0$$

$$\frac{5}{4}$$

There is no value of $$k$$ that fits the information.