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Q.
Find the value of $r$ if the coefficients of $(2r + 4)^{th}$ term and $(r - 2)^{th}$ term in the expansion of $(1 + x)^{18}$ are equal.
Binomial Theorem
Solution:
$T_{2r+4}=\,{}^{18}C_{2r+3}\,\left(x\right)^{2r+3}$ and
$T_{r-2} = \,{}^{18}C_{r-3}\,x^{r-3}$
We are given that $\,{}^{18}C_{2r+3} = \,{}^{18}C_{r-3}$
$\Rightarrow $ either $2r + 3 = r - 3$ or
$ 2r + 3 = 18 - \left(r - 3\right)$
$\Rightarrow $ either $r = - 6$ or $3r = 18$
$\Rightarrow r=6 \,\,$ ($\because r$ cannot be $-ve$)