Recall that a cardinal $\kappa$ is $(\lambda,\infty)$-almost-strongly-compact if every $\kappa$-complete filter can be refined to a $\lambda$-complete ultrafilter. A cardinal $\mu$ has the tree property if every $\mu$-sized tree with $\mu$-small levels has a branch of length $\mu$. (If in addition $\mu$ is inaccessible then $\mu$ is weakly compact.)
Question: Can the following constellation occur?
$\mu$ -- weakly inaccessible with the tree property
$\kappa$ -- a $(\mu^+,\infty)$-strongly-compact cardinal
every regular $\nu \in [\mu, \kappa)$ -- has the tree property.
I suspect this would be too good to be true. But I don't know much --
for all I know, maybe almost strong compactness implies inaccessibility, in which case of course the answer is no. But I'm having trouble tracking down even that information.
If $\kappa$ can be taken to be $(\mu,\infty)$-strongly-compact, that might be good enough for what I need. Also it should suffice for only the successor cardinals in $(\mu,\kappa)$ to have the tree property.
I apologize for the repeated changes to the question.