SG. 4 Let $S_{1}, S_{2}, \ldots, S_{10}$ be the first ten terms of an A.P., which consists of positive integers. If $\mathrm{S}_{1}+\mathrm{S}_{2}+\ldots+\mathrm{S}_{10}=55$ and $\left(\mathrm{S}_{10}-\mathrm{S}_{8}\right)+\left(\mathrm{S}_{9}-\mathrm{S}_{7}\right)+\ldots+\left(\mathrm{S}_{3}-\mathrm{S}_{1}\right)=d$, find $d$.
Reference: 1997 FG5.4
Let the general term be $S_{n}=a+(n-1) t$
$$
\begin{array}{l}
\frac{10}{2}[2 a+(10-1) t]=55 \\
\Rightarrow 2 a+9 t=11
\end{array}
$$
$\because a, t$ are positive integers, $a=1, t=1$
$$
\begin{aligned}
d & =\left(\mathrm{S}_{10}-\mathrm{S}_{8}\right)+\left(\mathrm{S}_{9}-\mathrm{S}_{7}\right)+\ldots+\left(\mathrm{S}_{3}-\mathrm{S}_{1}\right) \\
& =[a+9 t-(a+7 t)]+[a+8 t-(a+6 t)]+\ldots+(a+2 t-a) \\
d & =2 t+2 t+2 t+2 t+2 t+2 t+2 t+2 t \\
& =16 t=16
\end{aligned}
$$