If the coefficients of x 7 in ( a x 2 + 1 2 b x ) 11 and x - 7 in ( a x - 1 3 b x 2 ) 11 are equal, then [2023]

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Solution

Q.
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 [2023]

1 64ab = 243

2 729ab = 32

3 32ab = 729

4 243ab = 64

Ans.
(2)

We have, $r^{t h}$ term in ${(a x^{2} + \frac{1}{2 b x})}^{11}$ is,

$T_{r + 1} = {}^{11}{C_{r}} {(a x^{2})}^{11 - r} {(\frac{1}{2 b x})}^{r} = {}^{11}{C_{r}} \frac{a^{11 - r}}{{(2 b)}^{r}} \cdot x^{22 - 3 r}$

For coefficient of $x^{7}$ , we have $22 - 3 r = 7 \Rightarrow r = 5$

$∴$ $T_{6} = {}^{11}{C_{5}} \frac{a^{6}}{2^{5} b^{5}} x^{7} ...(i)$

Similarly, $r^{t h}$ term in the second expansion is, $T_{r + 1} = {}^{11}{C_{r}} {(a x)}^{11 - r} {(\frac{- 1}{3 b x^{2}})}^{r} = {}^{11}{C_{r}} \frac{a^{11 - r}}{{(- 3 b)}^{r}} \cdot x^{11 - 3 r}$

For coefficient of $x^{- 7}$ , we have $11 - 3 r = - 7 \Rightarrow r = 6$

$∴$ $T_{7} = {}^{11}{C_{6}} \frac{a^{5}}{3^{6} \cdot b^{6}} x^{- 7} ...(ii)$

Now, ${}^{11}{C_{5}} \frac{a^{6}}{2^{5} b^{5}} = {}^{11}{C_{6}} \frac{a^{5}}{3^{6} b^{6}}$ $\Rightarrow 3^{6} a b = 2^{5} \Rightarrow 729 a b = 32$

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