10. Given a sequence of positive terms $\left\{a_{n}\right\}$ with the sum of the first $n$ terms being $S_{n}$. If $\left\{a_{n}\right\},\left\{\sqrt{S_{n}}\right\}$ are both arithmetic sequences with a common difference of $d$, then $S_{n}=$
$$
\begin{array}{l}
\sqrt{S_{n}}=\sqrt{a_{1}}+(n-1) d \Rightarrow S_{n}=a_{1}+(n-1)^{2} d^{2}+2 d \sqrt{a_{1}}(n-1) \\
\Rightarrow \frac{d}{2} n^{2}+\left(a_{1}-\frac{d}{2}\right) n=d^{2} n^{2}+\left(2 d \sqrt{a_{1}}-2 d^{2}\right) n+a_{1}+d^{2}-2 d \sqrt{a_{1}} \\
\Rightarrow\left\{\begin{array} { l } 
{ d ^ { 2 } = \frac { d } { 2 } , } \\
{ 2 d \sqrt { a _ { 1 } } - 2 d ^ { 2 } = a _ { 1 } - \frac { d } { 2 } , } \\
{ a _ { 1 } + d ^ { 2 } - 2 d \sqrt { a _ { 1 } } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
d=\frac{1}{2}, \\
a_{1}=\frac{1}{4}
\end{array} \Rightarrow a_{n}=\frac{2 n-1}{4} \Rightarrow S_{n}=\frac{n^{2}}{4} .\right.\right.
\end{array}
$$