If the 4^ th term in the expansion of ( x2- 2y3)^6 is -20 , then xy=

Correct answer: B. (b) Now T_3-1.=^6C_3 ( x2 )^6-3 ( -2y3 )^3 ⇒-20= ( -2027 )x^3y^3 ⇒x^3y^3=27 ⇒ xy= (27 )^ 13=3 ⇒ xy=3

JEE Previous Year Question

$If the 4^{th} term in the expansion of (\frac{x}{2} - \frac{2 y}{3})^{6} is -20 , then xy =$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

Correct answer: B

(b) Now T_{3 - 1.} =^{6} C_{3} \times (\frac{x}{2})^{6 - 3} \times (\frac{- 2 y}{3})^{3} \Rightarrow - 20 = (\frac{- 20}{27}) x^{3} y^{3} \Rightarrow x^{3} y^{3} = 27 \Rightarrow xy = (27)^{\frac{1}{3}} = 3 \Rightarrow xy = 3

$The coefficient of x^{3} in the expansion of (1 - x)^{3/2}, (| x | < 1) is$
$If A denotes the sum of all the coefficients in the expansion of (1 - 3 x + 10 x^{2})^{n} and B denotes the sum of all the coe\dots$
$If the ratio of the 7th term from the beginning to the 7 th term from the end in the expansion of\dots$
$The remainder when ((64)^{(64)})^{(64)} is divided by 7 is equal to$
$Let m = (9 n^{2} + 54 n + 80) (9 n^{2} + 45 n + 54) (9 n^{2} + 36 n + 35) . The greatest positive integer which divides m, for all positi\dots$