What is the mass of the third element if element A has atomic mass 14 and middle element B has atomic mass 31 in Döbereiner’s triads?

The mass of the third element if element A has atomic mass $14$ and middle element B has atomic mass $31$ in Döbereiner’s triads is $48$.
Döbereiner’S Triads are groups of three elements with similar chemical and physical properties. In every triad, the atomic weight of the central element was half of the sum of the atomic weight of the other two elements.

We can see that the triad we have here is A, B, C. Here, B is the middle element. So, according to the law of triads, the mean of the atomic masses of A and C is expected to be equal to the atomic mass of B.
$\mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{B}=\frac{\mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{A}+\mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{C}}{2}$
$\text{⇒31 = }\frac{14+\mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{C}}{2}$
$\Rightarrow 31\times 2=14+ \mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{C}$
$\Rightarrow 62=14+ \mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{C}$
Hence, the atomic mass of $C$ is $48$.