We are given a sequence $Q$ with the following properties:
1. It consists of $\(\mathbf{1 5}\)$ numbers arranged in ascending order.
2. The first term $\(\left(a_1\right)$ is $\mathbf{2 5 .}\)$
3. For the first $\(\mathbf{1 4}\)$ terms, the ratio of a term to the next term is a fixed constant (let's denote this common ratio by $r$ ).
4. The last term ( $\(a_{15}\)$ ) is four times the first term, i.e., $\(a_{15}=4 \times 25=100\)$.

We need to find the 8th term $\(\left(a_8\right)\)$ of the sequence.
Step 1: Understand the Sequence Type
Since the ratio of consecutive terms is constant, this is a geometric sequence (at least for the first 14 terms). However, the last term is given separately, so we need to verify if the entire sequence follows the same pattern.

Step 2: Express the Terms of the Sequence
In a geometric sequence, the $n$-th term is given by:

$$
\(a_n=a_1 \times r^{n-1}\)
$$


Given:
- $\(a_1=25\)$
- $\(a_{15}=100\)$

But the sequence is geometric only for the first $\(\mathbf{1 4}\)$ terms. The $\(\mathbf{1 5 t h}\)$ term is given separately, so we can express the first 14 terms as:

$$
\(a_n=25 \times r^{n-1}, \quad \text { for } n=1 \text { to } 14\)
$$


The 15th term is:

$$
\(a_{15}=100\)
$$


Step 3: Find the Common Ratio $r$
The 14th term is:

$$
\(a_{14}=25 \times r^{13}\)
$$


The 15 th term is given as 100 , but it's not necessarily $a_{14} \times r$. Instead, the problem states that the ratio is fixed for the first 14 terms, and the last term is independently given.

However, since the sequence is ascending and the ratio is fixed for the first 14 terms, we can assume that the 15th term also follows the same ratio (unless specified otherwise). But the problem says the ratio is fixed for the first $\mathbf{1 4}$ terms, so the 15th term might not follow the same ratio.

But the problem also states that the last term is four times the first term, which is 100 . So, we can consider the entire sequence as geometric with 15 terms, where:

$$
\(a_{15}=25 \times r^{14}=100\)
$$


Solving for $r$ :

$$
\(\begin{aligned}
& r^{14}=\frac{100}{25}=4 \\
& r=4^{\frac{1 }{ 14}}=2^{\frac{1 }{ 7}}
\end{aligned}\)
$$


Step 4: Find the 8th Term
Using the geometric sequence formula:

$$
\(a_8=25 \times r^7=25 \times\left(2^{\frac{1 }{ 7}}\right)^7=25 \times 2^1=50\)
$$


Verification
Let's verify the common ratio and the 15th term:
- Common ratio $\(r=2^{\frac{1 }{ 7}}\)$.
- 15th term:

$$
\(a_{15}=25 \times\left(2^{\frac{1 }{ 7}}\right)^{14}=25 \times 2^2=100\)
$$


This matches the given condition.
Final Answer
The 8th term of the sequence is:

$$
\(50\)
$$