Given that, $ f (x+2) = f(x) + f(x+1) \quad \longrightarrow (1) $

And, $f(11) = 91 , f(15) =617 $

Let, $ f(10) $ be $ k.$

Now, from equation $(1) $

$ f(12) = f(10) + f(11) $

$ \Rightarrow f(12) = k + 91 \quad \longrightarrow (2) $

$ \Rightarrow f(13) = f (11) + f (12)$

$ \Rightarrow f(13) = 91 + k + 91 = k + 182 \quad \longrightarrow (3) $

$ \Rightarrow f(14) = f(12) +f(13) $

$ \Rightarrow f(14) =k + 91 + k + 182 $

$ \Rightarrow f(14) = 2k + 273 \quad \longrightarrow (4) $

$ \Rightarrow f(15) = f(13) + f(14) $

$ \Rightarrow 617 = k + 182 + 2k + 273 $

$ \Rightarrow 3k = 617 – 455 $

$ \Rightarrow 3k = 162 $

$ \Rightarrow k = \frac{162}{3}$

$ \Rightarrow \boxed{k = 54} $

$ \therefore \boxed {f(10) = k = 54} $

Correct Answer $: 54 $