If the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of 21/3+3−1/3n  is  1/6 then the value of n is

Let  a=7th term from the beginning=T7=T6+1=nC621/3n−63−1/36and let b=7th term from the end=(n+2−7) th or (n−5) th  term from the beginning=nCn−621/3n−(n−6)3−1/3n−6=nC621/363−1/3n−6we are given ab=1621/3n−63−1/3621/363−1/3n−6=16⇒21/33−1/3n−12=16⇒6(n−12)/3=6−1⇒n−12=−3⇒n=9.