If the coefficients of 2^nd , 3^rd and 4^th terms in the expansion of (1+x)ⁿ are in A.P. then find the value of n. Hence find the middle terms

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Please find below the solution to the asked query:



$$\begin{gathered} {\mathrm{n}}^2-2\mathrm{n}-7\mathrm{n}+14=0 \\ \Rightarrow \mathrm{n}\left(\mathrm{n}-2\right)-7\left(\mathrm{n}-2\right)=0 \\ \Rightarrow \left(\mathrm{n}-2\right)\left(\mathrm{n}-7\right)=0 \\ \mathrm{n}\ne 2 \\ \Rightarrow \mathrm{n}=7 \\ \mathrm{Middle} \mathrm{terms} \mathrm{will} \mathrm{be} \frac{7+1}{2}=4\mathrm{th} \mathrm{and} \frac{7+3}{2}=5\mathrm{th} \mathrm{terms} \mathrm{in} {\left(1+\mathrm{x}\right)}^7 \\ {\mathrm{T}}_4={\mathrm{T}}_{3+1}{=}^7{\mathrm{C}}_3 {\mathrm{x}}^3 \\ {\mathrm{T}}_5={\mathrm{T}}_{4+1}{=}^7{\mathrm{C}}_4 {\mathrm{x}}^4 \end{gathered}$$

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