If the coefficients of x7 in (ax2+12bx)11 and x−7 in (ax−13bx2)11 are equal then

Tr+1=11Cr(ax2)11−r(12bx)r ∴ 22−3r=7⇒r=5 Tr+1=11Cr(ax)11−r(−13bx2)r 11−3r=−7⇒r=6Equating the terms, we get ⇒ a32b5=1729b6⇒729ab=32

If the coefficients of x7 in (ax2+12bx)11 and x−7 in (ax−13bx2)11 are equal then

Tr+1=11Cr(ax2)11−r(12bx)r ∴ 22−3r=7⇒r=5 Tr+1=11Cr(ax)11−r(−13bx2)r 11−3r=−7⇒r=6Equating the terms, we get ⇒ a32b5=1729b6⇒729ab=32

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

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$729 a b = 32$

$243 a b = 64$

$32 a b = 729$

$64 a b = 243$

answer is C.

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Detailed Solution

$T_{r + 1} = 11_{C_{r}} {(a x^{2})}^{11 - r} {(\frac{1}{2 b x})}^{r}$
$∴ 22 - 3 r = 7 \Rightarrow r = 5$
$T_{r + 1} = 11_{C_{r}} {(a x)}^{11 - r} {(\frac{- 1}{3 b x^{2}})}^{r}$
$11 - 3 r = - 7 \Rightarrow r = 6$
Equating the terms, we get
$\Rightarrow \frac{a}{32 b^{5}} = \frac{1}{729 b^{6}} \Rightarrow 729 a b = 32$