If the coefficients of $$x^4$$, $$x^5$$ and $$x^6$$ in the expansion of $$(1 + x)^n$$ are in the arithmetic progression, then the maximum value of $$n$$ is:

The coefficient of $$x^k$$ in the expansion of $$(1+x)^n$$ is $${}^{n}C_{k}$$ (the binomial coefficient).

Given that the coefficients of $$x^{4},\,x^{5},\,x^{6}$$ are in arithmetic progression (A.P.), they must satisfy the A.P. condition
$$2\;{}^{n}C_{5}=\,{}^{n}C_{4}+{}^{n}C_{6}\quad -(1)$$

To simplify $$-(1)$$, divide every term by $${}^{n}C_{5}$$:

$$2=\frac{{}^{n}C_{4}}{{}^{n}C_{5}}+\frac{{}^{n}C_{6}}{{}^{n}C_{5}}\quad -(2)$$

Evaluate each ratio separately.
Using $$\displaystyle {}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}$$:

$$\frac{{}^{n}C_{4}}{{}^{n}C_{5}} =\frac{\dfrac{n!}{4!\,(n-4)!}}{\dfrac{n!}{5!\,(n-5)!}} =\frac{5!\,(n-5)!}{4!\,(n-4)!} =5\;\frac{(n-5)!}{(n-4)!} =\frac{5}{\,n-4\,} \quad -(3)$$

$$\frac{{}^{n}C_{6}}{{}^{n}C_{5}} =\frac{\dfrac{n!}{6!\,(n-6)!}}{\dfrac{n!}{5!\,(n-5)!}} =\frac{5!\,(n-5)!}{6!\,(n-6)!} =\frac{(n-5)!}{6\,(n-6)!} =\frac{\,n-5\,}{6}\quad -(4)$$

Substitute $$(3)$$ and $$(4)$$ into $$(2)$$:

$$2=\frac{5}{n-4}+\frac{n-5}{6}$$

Clear denominators by multiplying through by $$6(n-4)$$:

$$12(n-4)=30+(n-5)(n-4)$$

Simplify both sides:

Left side: $$12n-48$$
Right side: $$(n-5)(n-4)=n^{2}-9n+20$$, so
$$30+(n^{2}-9n+20)=n^{2}-9n+50$$

Set the two sides equal:

$$12n-48=n^{2}-9n+50$$

Bring all terms to the right:

$$0=n^{2}-9n+50-12n+48$$
$$\Rightarrow n^{2}-21n+98=0 \quad -(5)$$

Factor or use the quadratic formula on $$(5)$$.
Discriminant: $$\Delta=21^{2}-4\cdot98=441-392=49=7^{2}$$

Hence

$$n=\frac{21\pm7}{2}\;\Longrightarrow\; n_{1}=7,\; n_{2}=14$$

The question asks for the maximum possible integer $$n$$, so

Answer: $$n=14$$ (Option D).