If the constant term in the expansion of ( 3 5 x + 2 x 5 3 ) 12 , ? x ? 0 , is × 2 8 × 3 5 , then 25 is equal to: [2024]

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Solution

Q.
If the constant term in the expansion of ${(\frac{\sqrt[5]{3}}{x} + \frac{2 x}{\sqrt[3]{5}})}^{12}, x \neq 0,$ is $α \times 2^{8} \times \sqrt[5]{3},$ then $25 α$ is equal to: [2024]

1 639

2 742

3 724

4 693

Ans.
(4)

$T_{r + 1} = {}^{12}{C_{r}} {(\frac{3^{1 / 5}}{x})}^{12 - r} {(\frac{2 x}{5^{1 / 3}})}^{r}$

$= {}^{12}C_{r} (\frac{3^{\frac{12 - r}{5}}}{5^{r / 3}}) \cdot 2^{r} \cdot x^{2 r - 12}$

For constant term, $2 r - 12 = 0$

$\Rightarrow r = 6$

$∴$ Constant term = $\frac{{}^{12}{C_{6} \cdot} 3^{6 / 5} \cdot 2^{6}}{5^{2}}$

$= 924 \times \frac{3^{1} 3^{1 / 5} \cdot 2^{6}}{25} = \frac{11 \times 9 \times 7 \times 2^{8} \times 3^{1 / 5}}{25}$

$\Rightarrow α = \frac{11 \times 9 \times 7}{25} \Rightarrow 25 α = 11 \times 9 \times 7 = 693$

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