$$\frac{2}{\frac{2\Delta}{b}}=\frac{1}{\frac{2\Delta}{a}}+\frac{1}{\frac{2\Delta}{c}}$$ $$2b=a+c$$ Thus a,b and c are in an arithmetic progression. But the answer is $\sin A,\sin B,\sin c$ are in AP.
I understand that both are equivalent, but just needed to make sure, is it possible that one is in AP and the other is not, or is the answer wrong?
(The answer very specifically slashed out a,b,c being in AP, which is why I am asking it here to confirm )
By the sine rule, $$ \boxed {\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = K}$$ for some $K > 0$.
Thus, if $a,b,c$ are in arithmetic progression with common difference $d$, so are $\sin A , \sin B , \sin C$ with common difference $\frac dK$.