The 7th term of `log_(e)(5//4)` is

The nth term of log_(e)(4//3) is

Observe the following lists. {:("LIST - I","LIST - II"),((A) 8^(th)" term of log"_(e)^(2) " is",(1)log((n+1)/(n-1))),((B)7^(th)" term of log"_(e )(5//4)" is",(2)(-1)/(8)),((C )n^(th)" term of log"_(e )(3//2)" is",(3)(1)/(7.4^(7))),((D)(2)/(n^(2)+1)+(1)/(3)((2n)/(n^(2)+1))^(3)+.....,(4)( (-1)^(n-1))/(n.2^(n))):} The correct match for List-I from List-II is

n^(th) term of log_(e )(6//5) is

n^(th) term of log_(e )(6//5) is

The 8^(th) term of log_(e )2 is

The 8^(th) term of log_(e )2 is

If a, b, c are the fifth terms of log_(e)(3//2),log_(e)(4//3),log_(e)(6//5) then the ascending order of a, b, c is

The 4^(th) term of "log"_(e )(3)/(2) is

The 4^(th) term of "log"_(e )(3)/(2) is