The question appears to be in Persian, and it translates to:

"Among the numbers -5 and $\frac{35}{4}$, we've inserted 4 arithmetic means. Which of these is the largest number among these 4 arithmetic means?"

The answer choices provided are:

1. $\frac{29}{4}$
2. $-5$
3. $\frac{23}{4}$
4. $\frac{13}{2}$

To solve this:

1. Identify the Terms in the Sequence: We have the first term $a = -5$ and the last term $l = \frac{35}{4}$, with 4 arithmetic means in between. Therefore, there are 6 terms in total, so this is an arithmetic sequence with 6 terms.

2. Determine the Common Difference: For an arithmetic sequence, the common difference $d$ can be found by the formula: $d = \frac{l - a}{n - 1}$ where $n$ is the total number of terms. Here, $n = 6$, so: $d = \frac{\frac{35}{4} - (-5)}{6 - 1} = \frac{\frac{35}{4} + 5}{5}$ Converting $-5$ to quarters: $d = \frac{\frac{35}{4} + \frac{20}{4}}{5} = \frac{\frac{55}{4}}{5} = \frac{55}{20} = \frac{11}{4}$

3. Calculate the Sequence Terms: The terms of the sequence are:

   • First term: $a = -5$
   • Second term: $a + d = -5 + \frac{11}{4} = -\frac{20}{4} + \frac{11}{4} = -\frac{9}{4}$
   • Third term: $a + 2d = -5 + 2 \times \frac{11}{4} = -5 + \frac{22}{4} = -\frac{20}{4} + \frac{22}{4} = \frac{2}{4} = \frac{1}{2}$
   • Fourth term: $a + 3d = -5 + 3 \times \frac{11}{4} = -5 + \frac{33}{4} = -\frac{20}{4} + \frac{33}{4} = \frac{13}{4}$
   • Fifth term: $a + 4d = -5 + 4 \times \frac{11}{4} = -5 + \frac{44}{4} = -\frac{20}{4} + \frac{44}{4} = \frac{24}{4} = 6$
   • Sixth term (last term): $l = \frac{35}{4}$

4. Identify the Largest Term Among the Means: Among the 4 arithmetic means, the largest value is the fifth term: $\frac{13}{4} = 6$