1. In the expansion of $(1+x)^n$, the coefficient of $x^r$ is $\binom{n}{r}.$ So the given coefficients are: $\binom{n}{4},\quad \binom{n}{5},\quad \binom{n}{6}.$

2. Since these are in arithmetic progression (A.P.), the middle term is the average of the other two: $2\binom{n}{5}=\binom{n}{4}+\binom{n}{6}.$

3. Use relations between consecutive binomial coefficients: $\binom{n}{5}=\binom{n}{4}\cdot \frac{n-4}{5},$ $\binom{n}{6}=\binom{n}{5}\cdot \frac{n-5}{6}=\binom{n}{4}\cdot \frac{(n-4)(n-5)}{30}.$

Substitute into the A.P. condition: $2\left[\binom{n}{4}\cdot \frac{n-4}{5}\right]=\binom{n}{4}+\binom{n}{4}\cdot \frac{(n-4)(n-5)}{30}.$

4. Since $n\ge 6$ for all these coefficients to exist and $\binom{n}{4}\neq 0$, divide throughout by $\binom{n}{4}$: $\frac{2(n-4)}{5}=1+\frac{(n-4)(n-5)}{30}.$

Multiply by $30$: $12(n-4)=30+(n-4)(n-5).$

Expand: $12n-48=30+n^2-9n+20.$ $12n-48=n^2-9n+50.$

Bring all terms to one side: $n^2-21n+98=0.$

Factorize: $(n-7)(n-14)=0.$

Thus, $n=7 \quad \text{or} \quad n=14.$

5. Therefore, the maximum value of $n$ is $\boxed{14}.$

6. Checking options:

• A: $28$ ❌
• B: $21$ ❌
• C: $7$ ❌
• D: $14$ ✅

Hence, the correct option is D.