The positive value of a such that the coeffic | vedclass

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(A) The general term $T_{r+1}$ in the expansion of $(x^2 + \frac{a}{x^3})^{10}$ is given by:$T_{r+1} = {}_{10}C_r (x^2)^{10-r} (\frac{a}{x^3})^r$$T_{r

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$\frac{1}{2\sqrt{3}}$

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(A) The general term $T_{r+1}$ in the expansion of $(x^2 + \frac{a}{x^3})^{10}$ is given by:$T_{r+1} = {}_{10}C_r (x^2)^{10-r} (\frac{a}{x^3})^r$$T_{r+1} = {}_{10}C_r x^{20-2r} \cdot a^r \cdot x^{-3r}$$T_{r+1} = {}_{10}C_r \cdot a^r \cdot x^{20-5r}$For the coefficient of $x^5$,set $20-5r = 5$ $\Rightarrow 5r = 15$ $\Rightarrow r = 3$. The coefficient is ${}_{10}C_3 a^3 = 120a^3$.For the coefficient of $x^{15}$,set $20-5r = 15$ $\Rightarrow 5r = 5$ $\Rightarrow r = 1$. The coefficient is ${}_{10}C_1 a^1 = 10a$.Equating the coefficients: $120a^3 = 10a$.Since $a > 0$,we can divide by $10a$: $12a^2 = 1 \Rightarrow a^2 = \frac{1}{12}$.Thus,$a = \sqrt{\frac{1}{12}} = \frac{1}{2\sqrt{3}}$.

The positive value of $a$ such that the coefficient of $x^5$ is equal to that of $x^{15}$ in the expansion of $(x^2 + \frac{a}{x^3})^{10}$ is

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