The Total solution for NCERT class 6-12
Find the 7th term from the end in the expansion of (2x2┬атАУ3/2x)┬а8.
Given:
(2x2┬атАУ3/2x)┬а8
Let Tr+1┬аbethe┬а4th term from the end of the given expression.
Then, Tr+1┬аis(9 тИТ 7 + 1)th term, i.e., 3rd term, from the beginning.
T3┬а= T2+1
тИ┤ The 7th┬аtermfrom the end is 4032 x10.