If the \(5^{\text {th }}, 6^{\text {th }}\) and \(7^{\text {th }}\) term of the binomial expansion of \(\left(1+x^2\righ…
If the \(5^{\text {th }}, 6^{\text {th }}\) and \(7^{\text {th }}\) term of the binomial expansion of \(\left(1+x^2\right)^{n+4}\) are in A.P. Then the greatest binomial coefficient in the expansion of \(\left(1+x^2\right)^{n+4}\) is(\(\mathrm{n}\neq 10\)) (24 Jan, Shift I, Memory Based)
35
\({T}_{5},{T}_{6}\&{T}_{7}areinA.P.\\ 2{T}_{6}={T}_{5}+{T}_{7}\)
\(2 \times{ }^{n+4} C_5={ }^{n+4} C_4+{}^{n+4}C_{6}\)
\(\Rightarrow \quad 2 \times \frac{(n+4)!}{5! \times (n-1)!}=\frac{(n+4)!}{4! \times n!}+\frac{(n+4)!}{6! \times (n-2)!}\)
\(\begin{aligned}& \Rightarrow \frac{2 }{5(n-1)}=\frac{1}{n(n-1)}+\frac{1}{6 \times 5} \\& \Rightarrow \frac{2}{5(n-1)}=\frac{1}{n(n-1)}+\frac{1}{30} \\& \Rightarrow 2 n=5+\frac{n(n-1)}{6}\end{aligned}\)
\(\begin{aligned}&\begin{array}{r}\quad 12 n=30+n^2-n \\\Rightarrow \quad n^2-13 n+30=0 \\\Rightarrow \quad(n-10)(n-3)=0 \\\quad n=10,(3) \\\end{array}\\&=\text { greatest binomial coeff }\\&={}^7 C_3=\frac{7!}{3!.4!}=7 \times 5=35\end{aligned}\)
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