Let a and b be two non-zero real numbers. If the coefficient of x 5 in the expansion of ( a x 2 + 70 27 b x ) 4 is equal to the coefficient of x - 5 in the expansion of ( a x - 1 b x 2 ) 7 , then the value of 2 b is [

Question

Let $a$ and $b$ be two non-zero real numbers. If the coefficient of $x^{5}$ in the expansion of ${(a x^{2} + \frac{70}{27 b x})}^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{7},$ then the value of $2 b$ is [2023]

Smart Achievers · Accepted Answer

(3)

$Given expansion {(a x^{2} + \frac{70}{27 b x})}^{4}$

$T_{r + 1} = {}^{4}C_{r} {(a x^{2})}^{4 - r} {(\frac{70}{27 b x})}^{r}$

$= {}^{4}C_{r} a^{4 - r} {(\frac{70}{27 b})}^{r} . x^{8 - 3 r}$

$Here, 8 - 3 r = 5 \Rightarrow r = 1$

$So, coefficient of x^{5} = {}^{4}C_{1} a^{3} \cdot \frac{70}{27 b}$

$For expansion {(a x - \frac{1}{b x^{2}})}^{7}$

$T_{r + 1} = {}^{7}C_{r} {(a x)}^{7 - r} {(\frac{- 1}{b x^{2}})}^{r} = {}^{7}C_{r} a^{7 - r} {(\frac{- 1}{b})}^{r} x^{7 - 3 r}$

$Here, 7 - 3 r = - 5 \Rightarrow r = 4$

$So, coefficient of x^{- 5} = {}^{7}C_{4} a^{3} {(\frac{- 1}{b})}^{3}$

$A.T.Q., {}^{4}C_{1} a^{3} \cdot \frac{70}{27 b} = {}^{7}C_{4} a^{3} \cdot \frac{- 1}{b^{3}}$

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

Solution

Q.
Let $a$ and $b$ be two non-zero real numbers. If the coefficient of $x^{5}$ in the expansion of ${(a x^{2} + \frac{70}{27 b x})}^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{7},$ then the value of $2 b$ is [2023]

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Solution

Q. Let a and b be two non-zero real numbers. If the coefficient of x5 in the expansion of (ax2+7027bx)4 is equal to the coefficient of x-5 in the expansion of (ax-1bx2)7, then the value of 2b is [2023]

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Q.
Let $a$ and $b$ be two non-zero real numbers. If the coefficient of $x^{5}$ in the expansion of ${(a x^{2} + \frac{70}{27 b x})}^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{7},$ then the value of $2 b$ is [2023]