If the term independent of $x$ in the expansion of $\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}$ is 105 , then $\mathrm{a}^2$ is equal to :

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Aicharya · Accepted Answer

$\left(\sqrt{\mathrm{a}} \mathrm{x}^2+\frac{1}{2 \mathrm{x}^3}\right)^{10}$ General term $={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\sqrt{\mathrm{a}} \mathrm{x}^2\right)^{10-\mathrm{r}}\left(\frac{1}{2 \mathrm{x}^3}\right)^{\mathrm{r}}$ $\begin{aligned} & 20-2 r-3 r=0 \\ & \mathrm{r}=4 \\ & { }^{10} \mathrm{C}_4 \mathrm{a}^3 \cdot \frac{1}{16}=105 \\ & \mathrm{a}^3=8 \\ & \mathrm{a}^2=4\end{aligned}$
SECTION - B

If the term independent of $x$ in the expansion of $\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}$…

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