A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of 213+12(3)1310 is

# A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of

1. A

$1:2\left(6{\right)}^{\frac{1}{3}}$

2. B

$1:4\left(16{\right)}^{\frac{1}{3}}$

3. C

$4\left(36{\right)}^{\frac{1}{3}}:1$

4. D

$2\left(36{\right)}^{\frac{1}{3}}:1$

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### Solution:

Since, rth term from the end in the expansion of a binomial (r +$\mathrm{\alpha }$)n is same as the (n - r +2)th term from the beginning in the expansion of same binomial.

$\begin{array}{r}\left[\because {\mathrm{T}}_{\mathrm{r}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}{\mathrm{a}}^{\mathrm{r}}\right]\\ \left[{\because }^{10}{\mathrm{C}}_{4}{=}^{10}{\mathrm{C}}_{6}\right]\end{array}$

$\begin{array}{r}=\frac{{2}^{6/3}{\left(2\left(3{\right)}^{1/3}\right)}^{6}}{{2}^{4/3}{\left(2\left(3{\right)}^{1/3}\right)}^{4}}\\ ={2}^{6/3-4/3}{\left(2\left(3{\right)}^{1/3}\right)}^{6-4}\\ ={2}^{23}\cdot {2}^{2}\cdot {3}^{23}\\ =4\left(6{\right)}^{23}\\ =4\left(36{\right)}^{1/3}\end{array}$

So, the required ratio is. $4\left(36{\right)}^{1/3}:1$

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