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Q.
If the ratio of the coefficient of third and fourth term in the expansion of $\left( x - \frac{1}{2x}\right)^{n}$ is $1 : 2$, then the value of $-n$ will be
Binomial Theorem
Solution:
$T_3 = \,{}^nC_2 (x)^{n-2} \left( -\frac{1}{2x}\right)^{2}$ and $T_4 = \,{}^nC_3(x)^{n-3} \left( -\frac{1}{2x}\right)^{3}$
But according to the condition,
$\frac{-n (n-1) \times 3 \times 2 \times 1 \times 8}{n(n-1) (n - 2) \times 2 \times 1 \times4} = \frac{1}{2}$
$ \Rightarrow n = - 10$