TS EAMCET 4 May 2018 Shift 1 Solved Paper

Consider the term.
(√[4]x−3+a√[4]x5)n
Its general term is given as,
Tr+1=Cr(

1

x 3 4

)n−rar(x

5

4

)r
=nCrarx−

3

4

n+

3

4

r+

5

4

r
Now the sum of all binomial coefficient is 2n and 200<2n<400
Therefore, n=8 .
Now for independent of x
−

3

4

n+2r=0
−

3

4

(8)+2r=0
2r=6
r=3
Therefore,
nCrar=448
8C3 a3=448
56a3=448
a=2