Mathematics - Binomial Theorem Question with Solution | TestHub

Given,

The coefficient of $x^5$ in ${\left(a{x}^2+\frac{70}{27bx}\right)}^4$ is equal to coefficient of $x^{-5}$ in ${\left(ax-\frac{1}{b{x}^2}\right)}^7$,

Now finding the coefficient of $x^5$ in ${\left(a{x}^2+\frac{70}{27bx}\right)}^4$ we get,

$T_{r+1}={}^{4}C_r{\left(a{x}^2\right)}^{4-r}·{\left(\frac{70}{27bx}\right)}^r$

$\Rightarrow T_{r+1}={}^{4}C_r{\left(a\right)}^{4-r}·{\left(\frac{70}{27b}\right)}^r·x^{8-2r-r}$

So, $8-2r-r=5\Rightarrow r=1$

So, $T_2={}^{4}C_1{a}^3·\frac{70}{27b}·x^5$

Now finding the coefficient of $x^{-5}$ in ${\left(ax-\frac{1}{b{x}^2}\right)}^7$ we get,

$T_{r+1}={}^{7}C_r{\left(ax\right)}^{7-r}·{\left(\frac{1}{b{x}^2}\right)}^r$

$\Rightarrow T_{r+1}={}^{7}C_r{\left(a\right)}^{7-r}·{\left(\frac{1}{b}\right)}^r·x^{7-r-2r}$

So, $7-r-2r=-5\Rightarrow r=4$

So, $T_5={}^{7}C_4{\left(a\right)}^3·{\left(\frac{1}{b}\right)}^4·x^{-5}$

Now equating the coefficient of $x^5 \& x^{-5}$ we get,

${}^{7}C_4{\left(a\right)}^3·{\left(\frac{1}{b}\right)}^4={}^{4}C_1{a}^3·\frac{70}{27b}$

$\Rightarrow 35a^3·{\left(\frac{1}{b}\right)}^4=4a^3·\frac{70}{27b}$

$\Rightarrow {\left(\frac{1}{b}\right)}^3=4\times \frac{2}{27}$

$\Rightarrow b=\frac{3}{2}\Rightarrow 2b=3$