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If the coefficient of $x^7$ in $\big(ax^2+\large\frac{1}{bx}\big)^{11}$ is equal to coefficient of $x^{-7}$ in $\big(ax-\large\frac{1}{bx^2}\big)^{11}$, then $a$ and $b$ satisfy the relation

$\begin{array}{1 1} a+b =1 \\ a-b =1 \\ ab=1 \\ \frac{a}{b}=1 \end{array}$

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  • $T_{r+1}$ in $(x+y)^n=^nC_rx^{n-r}.y^r$
$T_{r+1}$ in $\big(ax^2+\large\frac{1}{bx}\big)^{11}$$=^{11}C_r(ax^2)^{11-r}.(\large\frac{1}{bx})^r$
$=^{11}C_r.\large\frac{a^{11-r}}{b^r}$$.x^{22-3r}$
For coeff. of $x^7,$ $22-3r=7$
$\Rightarrow\:r=5$
$\therefore$ Coeff. of $x^7$ in $\big(ax^2+\large\frac{1}{bx}\big)^{11}$$=^{11}C_5.\large\frac{a^6}{b^5}$.....(i)
Similarly
$T_{r+1}$ in $\big(ax-\large\frac{1}{bx^2}\big)^{11}$$=(-1)^r.$$ ^{11}C_r(ax)^{11-r}.(\large\frac{1}{bx^2})^r$
$=(-1)^n$.$^{11}C_r.\large\frac{a^{11-r}}{b^r}$$.x^{11-3r}$
For coeff. of $x^{-7},$ $11-3r=-7$
$\Rightarrow\:r=6$
$\therefore$ Coeff. of $x^{-7}$ in $\big(ax-\large\frac{1}{bx^2}\big)^{11}$$=^{11}C_6.\large\frac{a^5}{b^6}$.........(ii)
Given :(i) = (ii)
$\Rightarrow\:^{11}C_5.\large\frac{a^6}{b^5}=^{11}C_6.\large\frac{a^5}{b^6}$
$\Rightarrow\:ab=1$
answered Sep 15, 2013 by rvidyagovindarajan_1
 

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