How do you write the nth term rule for the sequence #5/2,11/6,7/6,1/2,-1/6,...#?

1 Answer
Aug 22, 2016

Answer:

#n^(th)# term of the series is #(19-4n)/6#

Explanation:

In #{5/2,11/6,7/6,1/2,-1/6...}#, let us make denominator common. As GCD is #6#, we can write the series as

#{15/6,11/6,7/6,3/6,-1/6...}#.

We observe that numerators form the series

#{15,11,7,3,-1,...}# which form an arithmetic series with first term #a# as #15# and common difference #d# between a term annd its preceding term is #12-15=-4#. Hence, #n^th# term of this series is #a+(n-1)d# i.e. #15+(n-1)×(-4)=15-4n+4=19-4n#

Hence #n^(th)# term of the series is #(19-4n)/6#