For some $\mathrm{n}\ne 10$, let the coefficients of the 5^th, 6^th and 7^th terms in the binomial expansion of (1 + x)ⁿ⁺⁴ be in A.P. Then the largest coefficient in the expansion of (1 + x)ⁿ⁺⁴ is :  

$\begin{array}{l}(1+\mathrm{x}{)}^{\mathrm{n}+4}\\ { }^{\mathrm{n}+4}{\mathrm{C}}_4{,}^{\mathrm{n}+4}{{\mathrm{C}}_5}^{\mathrm{n}+4}{\mathrm{C}}_6,\to \text{ A.P. }\\ \Rightarrow 2{\times }^{\mathrm{n}+4}{\mathrm{C}}_5{=}^{\mathrm{n}+4}{\mathrm{C}}_4{+}^{\mathrm{n}+4}{\mathrm{C}}_6\\ \Rightarrow 4{\times }^{\mathrm{n}+4}{\mathrm{C}}_5=\left({ }^{\mathrm{n}+4}{\mathrm{C}}_4{+}^{\mathrm{n}+4}{\mathrm{C}}_5\right)+\left({ }^{\mathrm{n}+4}{\mathrm{C}}_5{+}^{\mathrm{n}+4}{\mathrm{C}}_6\right)\\ \Rightarrow 4{\times }^{\mathrm{n}+4}{\mathrm{C}}_5{=}^{\mathrm{n}+5}{\mathrm{C}}_5{+}^{\mathrm{n}+5}{\mathrm{C}}_6\\ \Rightarrow 4\times \frac{(\mathrm{n}+4)!}{5!.(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\cdot \mathrm{n}!}\\ \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6\mathrm{n}}\\ \Rightarrow {\mathrm{n}}^2+11\mathrm{n}+30=24\mathrm{n}\\ \Rightarrow {\mathrm{n}}^2-13\mathrm{n}+30=0\\ \Rightarrow \mathrm{n}=3,10\left(\text{ rejected }\right)\\ \because \mathrm{n}\ne 10\end{array}$
$\therefore$Largest binomial coefficient in expansion of (1 + x)⁷
$\begin{array}{l}(\because \mathrm{n}+4=7)\\ \text{ is coeff. of middle term }\\ {\Rightarrow }^7{\mathrm{C}}_4{=}^7{\mathrm{C}}_3=35\end{array}$