+1}={ }^4 \mathrm{C}_{\mathrm{r}}\left(\mathrm{a} \cdot \mathrm{x}^2\right)^{4-\mathrm{r}} \cdot\left(\frac{70}{27 \mathrm{bx}}\right)^{\mathrm{r}} \\\\ & ={ }^4 \mathrm{C}_{\mathrm{r}} \cdot \mathrm{a}^{4-\mathrm{r}} \cdot \frac{70^{\mathrm{r}}}{(27 b)^{\mathrm{r}}} \cdot x^{8-3 \mathrm{r}} \end{aligned} $$

$$ \begin{aligned} & \text { here } 8-3 r=5 \\\\ & 8-5=3 r \Rightarrow r=1 \\\\ & \therefore \text { coeff. }=4 \cdot a^3 \cdot \frac{70}{27 b} \end{aligned} $$

$$ \begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^7 \mathrm{C}_{\mathrm{r}}(\mathrm{ax})^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{bx}^2}\right)^{\mathrm{r}} \\\\ & ={ }^7 \mathrm{C}_{\mathrm{r}} \cdot a^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{~b}}\right)^{\mathrm{r}} \cdot \mathrm{x}^{7-3 \mathrm{r}} \\\\ & 7-3 \mathrm{r}=-5 \Rightarrow 12=3 \mathrm{r} \Rightarrow \mathrm{r}=4 \end{aligned} $$

$$ \begin{aligned} & \text { now } \frac{35 \mathrm{a}^3}{\mathrm{~b}^4}=\frac{280 \mathrm{a}^3}{27 \mathrm{~b}} \\\\ & \mathrm{~b}^3=\frac{35 \times 27}{280}=\mathrm{b}=\frac{3}{2} \Rightarrow 2 \mathrm{~b}=3 \end{aligned} $$19:T4a2,$$ \begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^4 \mathrm{C}_{\mathrm{r}}\left(\mathrm{a} \cdot \mathrm{x}^2\right)^{4-\mathrm{r}} \cdot\left(\frac{70}{27 \mathrm{bx}}\right)^{\mathrm{r}} \\\\ & ={ }^4 \mathrm{C}_{\mathrm{r}} \cdot \mathrm{a}^{4-\mathrm{r}} \cdot \frac{70^{\mathrm{r}}}{(27 b)^{\mathrm{r}}} \cdot x^{8-3 \mathrm{r}} \end{aligned} $$

$$ \begin{aligned} & \text { यहाँ } 8-3 r=5 \\\\ & 8-5=3 r \Rightarrow r=1 \\\\ & \therefore \text { गुणांक }=4 \cdot a^3 \cdot \frac{70}{27 b} \end{aligned} $$

$$ \begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^7 \mathrm{C}_{\mathrm{r}}(\mathrm{ax})^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{bx}^2}\right)^{\mathrm{r}} \\\\ & ={ }^7 \mathrm{C}_{\mathrm{r}} \cdot a^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{~b}}\right)^{\mathrm{r}} \cdot \mathrm{x}^{7-3 \mathrm{r}} \\\\ & 7-3 \mathrm{r}=-5 \Rightarrow 12=3 \mathrm{r} \Rightarrow \ma

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