To find the missing numbers in the sequence \(-\frac{3}{4}, \_, \_, \frac{1}{4}\), we can assume that the sequence is an arithmetic progression (AP), where the difference between consecutive terms is constant. 

Let the common difference be \(d\). The sequence can be represented as:

\[
-\frac{3}{4}, -\frac{3}{4} + d, -\frac{3}{4} + 2d, \frac{1}{4}
\]

We know the fourth term is \(\frac{1}{4}\). So, we can write:

\[
-\frac{3}{4} + 3d = \frac{1}{4}
\]

Now, solve for \(d\):

\[
3d = \frac{1}{4} + \frac{3}{4} = 1
\]

\[
d = \frac{1}{3}
\]

Now, substitute \(d = \frac{1}{3}\) back into the sequence to find the missing terms:

- The second term is \( -\frac{3}{4} + \frac{1}{3} \).
- The third term is \( -\frac{3}{4} + 2 \times \frac{1}{3} \).

To calculate these:

1. The second term:

\[
-\frac{3}{4} + \frac{1}{3} = -\frac{9}{12} + \frac{4}{12} = -\frac{5}{12}
\]

2. The third term:

\[
-\frac{3}{4} + \frac{2}{3} = -\frac{9}{12} + \frac{8}{12} = -\frac{1}{12}
\]

Thus, the complete sequence is:

\[
-\frac{3}{4}, -\frac{5}{12}, -\frac{1}{12}, \frac{1}{4}
\]

Would you like any more details or have any questions? Here are five related questions you might find interesting:

1. How do you find the common difference in an arithmetic sequence?
2. What is the formula for the nth term of an arithmetic progression?
3. How do you determine if a sequence is arithmetic or geometric?
4. Can the common difference in an arithmetic sequence be negative?
5. How do you find the sum of the first n terms of an arithmetic sequence?

**Tip:** When dealing with fractions in sequences, it's helpful to convert them to a common denominator for easier addition and subtraction.

Learn how to find the missing numbers in the arithmetic sequence -3/4, _, _, 1/4. This problem involves arithmetic progressions and the formula for the nth term of an arithmetic sequence. Suitable for students in Grades 9-12.

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