For what values of n, ⁿc_1, ⁿc₂ and ⁿc₃ will be in arithmetic progression?

Dear Student,

$$\begin{gathered} \mathrm{Given}, {}^{\mathrm{n}}\mathrm{C}_1, {}^{\mathrm{n}}\mathrm{C}_2 \mathrm{and} {}^{\mathrm{n}}\mathrm{C}_3 \mathrm{are} \mathrm{in} \mathrm{A}.\mathrm{P}. \mathrm{So}, \\ 2{}^{\mathrm{n}}\mathrm{C}_2={}^{\mathrm{n}}\mathrm{C}_1+{}^{\mathrm{n}}\mathrm{C}_3 \\ \Rightarrow 2\left(\frac{\mathrm{n}!}{2!\left(\mathrm{n}-2\right)!}\right)=\frac{\mathrm{n}!}{1!\left(\mathrm{n}-1\right)!}+\frac{\mathrm{n}!}{1!\left(\mathrm{n}-3\right)!} \\ \Rightarrow \frac{1}{\left(\mathrm{n}-2\right)!}=\frac{1}{\left(\mathrm{n}-1\right)!}+\frac{1}{\left(\mathrm{n}-3\right)!} \\ \Rightarrow \frac{1}{\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)!}=\frac{1}{\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)!}+\frac{1}{\left(\mathrm{n}-3\right)!} \\ \Rightarrow \frac{1}{\left(\mathrm{n}-2\right)}=\frac{1}{\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)}+1 \\ \Rightarrow 1=\frac{1}{\left(\mathrm{n}-2\right)}-\frac{1}{\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)} \\ \Rightarrow 1=\frac{\mathrm{n}-1-1}{\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)} \\ \Rightarrow 1=\frac{1}{\left(\mathrm{n}-1\right)} \\ \Rightarrow \mathrm{n}-1=1 \\ \Rightarrow \mathrm{n}=1+1 \\ \Rightarrow \mathbf{n}\mathbf{=}\mathbf{2}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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