The third term from the end in the expansion of 4x-3y-3y-2x 9

The correct option is B ^ 9C_7 3^5/2^3 y^5/x^5 r th term #160; from end is equal to (n r+2)th term from begining T_(9 3+2) = T_8 = ^9C_7 (4x3y)^9 7 ...

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Standard XII

Mathematics

Definition of a Determinant

The third ter...

Question

The third term from the end in the expansion of ${(\frac{4 x}{3 y} - \frac{3 y}{2 x})}^{9}$

^{9} C_{7} \frac{3^{5}}{2^{3}} \frac{y^{5}}{x^{5}}

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^{- 9} C_{7} \frac{3^{5}}{2^{3}} \frac{y^{5}}{x^{5}}

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^{9} C_{7} \frac{3^{5}}{2^{3}} \frac{y^{5}}{x^{3}}

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Solution

The correct option is B $^{- 9} C_{7} \frac{3^{5}}{2^{3}} \frac{y^{5}}{x^{5}}$
r th term from end is equal to (n-r+2)th term from begining

$T_{(} 9 - 3 + 2) = T_{8} =^{9} C_{7} (\frac{4 x}{3 y})^{9 - 7} (\frac{- 3 y}{2 x})^{7}$
$= -^{9} C_{7} \frac{3^{7 - 2}}{2^{7 - 4}} \frac{y^{7 - 2}}{x^{7} - 2}$
$= -^{9} C_{7} \frac{3^{5}}{2^{3}} \frac{y^{5}}{x^{5}}$

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The third term from the end in the expansion of (4x3y−3y2x)9

The correct option is B −9C73523y5x5r th term from end is equal to (n-r+2)th term from beginingT(9−3+2)=T8=9C7(4x3y)9−7(−3y2x)7 =−9C737−227−4y7−2x7−2 =−9C73523y5x5

The third term from the end in the expansion of ${(\frac{4 x}{3 y} - \frac{3 y}{2 x})}^{9}$