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Question

Find the 6th term of the expansion (y1/2+x1/3)n, if the value of coefficient of 3rd term from the end is 45

A
232y5/2x5/3
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B
252y5/2x5/3
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C
232y5/3x5/2
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D
252y5/3x5/2
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Solution

The correct option is B 252y5/2x5/3
Coefficient of 3rd term from the beginning is always equal to the coefficient of 3rdterm from the end.

Now,
Tr+1= nCr (y1/2)(nr)(x1/3)r
Coefficient of T3= nC2=n(n1)2

n(n1)2=45n2n90=0n210n+9n90=0(n10)(n+9)=0n=10 (nN)

T6=T5+1T6= 10C5(y1/2)105(x1/3)5T6=10×9×8×7×65×4×3×2×1×y5/2x5/3T6=252y5/2x5/3

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