10.7. For any positive integers $a, b (a>b>1)$, define the sequence $x_{n}=\frac{a^{n}-1}{b^{n}-1}(n=1,2, \cdots)$. It is known that the defined sequence does not have $d$ consecutive terms consisting of prime numbers. Find the minimum value of $d$.

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10.7. The minimum value of $d$ is 3.

When $a=4, b=2$, $x_{1}=3, x_{2}=5$ are prime numbers.
Therefore, the minimum value of $d$ is greater than 2.
Next, we prove: there cannot be three consecutive terms all being prime numbers.
In fact, a stronger conclusion can be proven:
For $n \geqslant 2, x_{n}$ and $x_{n+1}$ cannot both be prime numbers.
Assume, for contradiction, that they are prime numbers $p$ and $q$. Then,
$$
\begin{array}{l}
(a-1)\left(a^{n-1}+a^{n-2}+\cdots+1\right) \\
=p(b-1)\left(b^{n-1}+b^{n-2}+\cdots+1\right), \\
(a-1)\left(a^{n}+a^{n-1}+\cdots+1\right) \\
=q(b-1)\left(b^{n}+b^{n-1}+\cdots+1\right) .
\end{array}
$$

Since $a^{n-1}+a^{n-2}+\cdots+1$ and $a^{n}+a^{n-1}+\cdots+1$ are coprime, we have
$$
(b-1) \mid (a-1).
$$

Let $a-1=k(b-1)$.
Substituting into the first equation, we get
$$
\begin{array}{l}
k\left(a^{n-1}+a^{n-2}+\cdots+1\right) \\
=p\left(b^{n-1}+b^{n-2}+\cdots+1\right).
\end{array}
$$

Since $a>b$, we have $1<k<p$.
Thus, $k \mid \left(b^{n-1}+b^{n-2}+\cdots+1\right)$.
Similarly, we have
$$
k<q, k \mid \left(b^{n}+b^{n-1}+\cdots+1\right),
$$

which contradicts the fact that $b^{n-1}+b^{n-2}+\cdots+1$ and $b^{n}+b^{n-1}+\cdots+1$ are coprime.