Let $a_{1}, a_{2}, \ldots, a_{n}, a_{n+1}$ be a sequence of distinct elements forming an arithmetic sequence. Write the sum

$$
S_{n}=x^{a_{2}^{2}-a_{1}^{2}}+x^{a_{3}^{2}-a_{2}^{2}}+\ldots+x^{a_{n+1}^{2}-a_{n}^{2}}
$$

in a simplified form, if $x \neq 1$.
Let $a_{1}, a_{2}, \ldots, a_{n+1}$ be an arithmetic sequence with common difference $d$. Transforming the exponents, we get:

$$
a_{2}^{2}-a_{1}^{2}=\left(a_{2}-a_{1}\right)\left(a_{2}+a_{1}\right)=d\left(a_{2}+a_{1}\right)=d\left(2 a_{1}+d\right)=2 a_{1} d+d^{2}
$$

Similarly:

$$
\begin{gathered}
a_{3}^{2}-a_{2}^{2}=2 a_{1} d+3 d^{2} \\
a_{n+1}^{2}-a_{n}^{2}=2 a_{1} d+(2 n-1) d^{2}
\end{gathered}
$$

Thus:

$$
\begin{aligned}
& S_{n}=x^{2 a_{1} d+d^{2}}+x^{2 a_{1} d+3 d^{2}}+\ldots+x^{2 a_{1} d+(2 n-1) d^{2}}= \\
& \quad=x^{2 a_{1} d+d^{2}}\left(1+x^{2 d^{2}}+x^{4 d^{2}}+\ldots+x^{(2 n-2) d^{2}}\right)
\end{aligned}
$$

In the parentheses, there is a geometric series with $n$ terms and a common ratio of $x^{2 d^{2}}$. Since it is given that $x \neq 1$ and $d \neq 0$, the common ratio is not 1, and we can apply the formula for the sum of a geometric series, which gives:

$$
S_{n}=x^{2 a_{1} d+d^{2}} \cdot \frac{x^{2 n d^{2}}-1}{x^{2 d^{2}}-1}
$$