How do you write the rule for the nth term given #-1/2,-1/4,-1/6,-1/8,...#?

1 Answer
Jul 29, 2016

Answer:

#n^(th)# of the sequence #{-1/2,-1/4,-1/6,-1/8, ..........}# is #-1/(2n)#

Explanation:

It is observed in the sequence that while numerator is constant at #-1#, the difference between the denominator of a term of the sequence to the denominator of its preceding term is constant at #2#, as denominators are #{2, 4, 6, 8, ..........}#.

Clearly, it is not the sequence but its denominators, which are in arithmetic sequence #{2, 4, 6, 8, ..........}#, with first term as #2# and common difference at #2#.

As the #n^(th)# of a sequence whose first term is #a# and common difference is #d# is #a+(n-1)d#, #n^(th)# of the sequence #{2, 4, 6, 8, ..........}# is

#2+(n-1)xx2=2+2n-2=2n#.

Hence #n^(th)# of the sequence #{-1/2,-1/4,-1/6,-1/8, ..........}# is #-1/(2n)#