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)

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YESPYQ · Accepted Answer

Correct answer: b) 35. ${T}_{5},{T}_{6}&{T}_{7}areinA.P.\ 2{T}_{6}={T}_{5}+{T}_{7}$ $2 \times{ }^{n+4} C_5={ }^{n+4} C_4+{}^{n+4}C_{6}$ $\Rightarrow \quad 2 \times \frac{(n+4)!}{5! \times (n-1)!}=\frac{(n+4)!}{4! \times n!}+\frac{(n+4)!}{6! \times (n-2)!}$ $\begin{aligned} & \Rightarrow \frac{2 }{5(n-1)}=\frac{1}{n(n-1)}+\frac{1}{6 \times 5} \ & \Rightarrow \frac{2}{5(n-1)}=\frac{1}{n(n-1)}+\frac{1}{30} \ & \Rightarrow 2 n=5+\frac{n(n-1)}{6} \end{aligned}$ $\begin{aligned} &\begin{array}{r} \quad 12 n=30+n^2-n \ \Rightarrow \quad n^2-13 n+30=0 \ \Rightarrow \quad(n-10)(n-3)=0 \ \quad n=10,(3) \ \end{array}\ &=\text { greatest binomial coeff }\ &={}^7 C_3=\frac{7!}{3!.4!}=7 \times 5=35 \end{aligned}$

If the \(5^{\text {th }}, 6^{\text {th }}\) and \(7^{\text {th }}\) term of the binomial expansion of \(\left(1+x^2\righ…

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