The Total solution for NCERT class 6-12
Find the 8th term in the expansion of (x3/2 y1/2 –x1/2 y3/2)10.
Given:
(x3/2 y1/2 –x1/2 y3/2)10
Let us consider the 8th termas T8
So,
T8 = T7+1
= 10C7 (x3/2 y1/2)10-7 (-x1/2 y3/2)7
= -[10×9×8]/[3×2] x9/2 y3/2 (x7/2 y21/2)
= -120 x8y12
∴ The 8th termof the expression (x3/2 y1/2 – x1/2 y3/2)10 is-120 x8y12.