Differential calculus using Rolles theorem and mean value theorem

Question

Q- A function f is such that its second derivative is continuous on [a, a+h] and Derivable on (a,a+h). show that there exists a number $\theta$ between 0 and 1 such that- $$ f(a+h) - f(a) -\frac{h[f'(a) +f'(a+h)] }{2} + \frac{(h^3)f''(a+\theta h)}{12} = 0$$ As I was not getting the answer , i tried forcibly to get the answer considering last terms as $\frac{h^3(f'''(a+\theta h))}{12}$ and assuming $\phi (x) =f(x) +\frac{(a+h-x)[f'(x) +f'(a+h)] }{2}$ $$-(a+h-x)^3 A$$ where A is arbitrary constant ,determined using $$\phi(a)=\phi(a+h) $$ Using rolles theorem $\exists ( a+\theta'h) \in (a,a+h)$ where $0<\theta'<1$ such that $\phi'(a+\theta'h)=0 $
Again we have $\phi'(x) = \frac{[f'(x)-f'(a+h)]}{2}+\frac{(a+h-x)f''(x)}{2}$ $$+3(a+h-x)^2 A$$ Again, $\exists ( a+\theta h) \in (a+\theta' h,a+h)$ where $0<\theta'<\theta<1$
clearly, $\phi'(a+\theta'h)=0 =\phi'(a+h)$
By rolles theorem $$\phi''(a+\theta h)=0$$
after solving we get $A = \frac{f'''(a+\theta h)}{12}$ that give the result.
Doubt 1- firstly i have done this by modifying the question . Is it possible to solve without changing the question if yes , do tell what $\phi(x)$ should be assumed
Doubt 2- whether i have assumed the correct $\phi(x)$, after modifying , as in other examples there were no such term where (a+h) remains intact as in above assumed $\phi(x) $ i.e. the term f'(a+h) in it.