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7.Binomial Theorem
hard
For some $n \neq 10$, let the coefficients of the $5^{\text {th }}, 6^{\text {th }}$ and $7^{\text {th }}$ terms in the binomial expansion of $(1+x)^{\text {n+4 }}$ be in $A.P.$ Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is :
A$70$
B$35$
C$20$
D$10$
(JEE MAIN-2025)
Solution
$(1+x)^{n+4}$
${ }^{n+4} C_4,{ }^{n+4} C_5,{ }^{n+4} C_6, \rightarrow \text { A.P. }$
$\Rightarrow 2 \times{ }^{n+4} C_5={ }^{n+4} C_4+{ }^{n+4} C_6$
$\Rightarrow 4 \times{ }^{n+4} C_5=\left({ }^{n+4} C_4+{ }^{n+4} C_5\right)+\left({ }^{n+4} C_5+{ }^{n+4} C_6\right)$
$\Rightarrow 4 \times{ }^{n+4} C_5={ }^{n+5} C_5+{ }^{n+5} C_6$
$\Rightarrow 4 \times \frac{(n+4)!}{5!\cdot(n-1)!}=\frac{(n+6)!}{6!\cdot n!}$
$\Rightarrow 4=\frac{(n+6)(n+5)}{6 n}$
$\Rightarrow n^2+11 n+30=24 n$
$\Rightarrow n^2-13 n+30=0$
$\Rightarrow n=3,10(\text { rejected })$
$\because n \neq 10$
$\therefore$ Largest binomial coefficient in expansion of
$(1+x)^7$
$(\because n+4=7)$
is coeff. of middle term
$\Rightarrow{ }^7 C_4={ }^7 C_3=35$
$N.T.A. $
${ }^{n+4} C_4,{ }^{n+4} C_5,{ }^{n+4} C_6, \rightarrow \text { A.P. }$
$\Rightarrow 2 \times{ }^{n+4} C_5={ }^{n+4} C_4+{ }^{n+4} C_6$
$\Rightarrow 4 \times{ }^{n+4} C_5=\left({ }^{n+4} C_4+{ }^{n+4} C_5\right)+\left({ }^{n+4} C_5+{ }^{n+4} C_6\right)$
$\Rightarrow 4 \times{ }^{n+4} C_5={ }^{n+5} C_5+{ }^{n+5} C_6$
$\Rightarrow 4 \times \frac{(n+4)!}{5!\cdot(n-1)!}=\frac{(n+6)!}{6!\cdot n!}$
$\Rightarrow 4=\frac{(n+6)(n+5)}{6 n}$
$\Rightarrow n^2+11 n+30=24 n$
$\Rightarrow n^2-13 n+30=0$
$\Rightarrow n=3,10(\text { rejected })$
$\because n \neq 10$
$\therefore$ Largest binomial coefficient in expansion of
$(1+x)^7$
$(\because n+4=7)$
is coeff. of middle term
$\Rightarrow{ }^7 C_4={ }^7 C_3=35$
$N.T.A. $
Standard 11
Mathematics
