If the coefficients of x^7 in left( ax ^2+fra | vedclass

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Question

$\left(a x^2+\frac{1}{2 b x}\right)^{11}$

$T _{ r +1}={ }^{11} C _{ r }\left( ax ^2\right)^{11- r } \cdot\left(\frac{1}{2 bx }\right)^{ r }$

$={

Vedclass · Accepted Answer

$729 ab =32$

Vedclass · Answer

$\left(a x^2+\frac{1}{2 b x}ight)^{11}$

$T _{ r +1}={ }^{11} C _{ r }\left( ax ^2ight)^{11- r } \cdot\left(\frac{1}{2 bx }ight)^{ r }$

$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{1}{2 b }ight)^{ I } \cdot x ^{22-2 r - r }={ }^{11} C _{ r } a^{11- r } \cdot\left(\frac{1}{2 b }ight)^{ I } \cdot x ^{22-3 r }$

$	herefore 22-3 r=7$

$3 r =15$

$r=5$

Again $\left(a x-\frac{1}{3 b x^2}ight)^{11}$

$T _{ r +1}={ }^{11} C _{ r }( ax )^{11- r }\left(-\frac{1}{3 b x^2}ight)^{ r }$

$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{-1}{3 b }ight)^{ r } \cdot x ^{11- r -2 r }$ $	herefore 11-3 r=-7$

$3 r =18$

$r=6$

$	ext { Now, } \frac{{ }^{11} C _5 a ^6}{32 b^5}=\frac{{ }^{11} C _6 \cdot a ^5}{3^6 \cdot b ^6}$

$729 ab =32$

If the coefficients of $x^7$ in $\left( ax ^2+\frac{1}{2 bx }\right)^{11}$ and $x ^{-7}$ in $\left(a x-\frac{1}{3 b x^2}\right)^{11}$ are equal, then

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