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.