Let a and b be two nonzero real numbers. If t | vedclass

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Question

$\mathrm{T}_{r+1}={ }^4 \mathrm{C}_r\left(a \mathrm{x}^2\right)^{4-r} \cdot\left(\frac{70}{27 \mathrm{bx}}\right)^r$

$={ }^{+} \mathrm{C}_{\mathrm{

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\mathrm{T}_{r+1}={ }^4 \mathrm{C}_r\left(a \mathrm{x}^2ight)^{4-r} \cdot\left(\frac{70}{27 \mathrm{bx}}ight)^r$

$={ }^{+} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{a}^{4-	au} \cdot \frac{70^r}{(27 \mathrm{~b})^r} \mathrm{x}^{s-3 t}$

here $8-3 r=5$

$8-5=3 r \Rightarrow r=1$

$	herefore 	ext { coeff. }=4 a^3 \cdot \frac{70}{27 b}$

$T_{r+1}={ }^7 C_r(a x)^{7-r}\left(\frac{-1}{b x^2}ight)^r$

$={ }^7 C_r \cdot a^{7-r}\left(\frac{-1}{b}ight)^r \cdot x^{7-3 r}$

$7-3 r=-5 \Rightarrow 12=3 r \Rightarrow r=4$

coeff. : ${ }^7 C_4 \cdot a^3 \cdot\left(\frac{-1}{b}ight)^4=\frac{35 a^3}{b^4}$

now $\frac{35 a^5}{b^4}=\frac{280 a^3}{27 b}$

$b^3=\frac{35 	imes 27}{280}=b=\frac{3}{2} \Rightarrow 2 b=3$

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $\left(a x^2+\frac{70}{27 b x}\right)^4$ is equal to the coefficient of $x^{-5}$ is equal to the coefficient of $\left(a x-\frac{1}{b x^2}\right)^7$, then the value of $2 b$ is

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