How do you find #a_6# for the geometric sequence #540, 90, 15,...#?

1 Answer
Oct 23, 2016

Answer:

The sixth term #a_6# is #5/72#

Explanation:

In a geometric series, whose first term is #a_1# and common ratio is #r#, #n^(th)# term #a_n# is given by #a_1xxr^(n-1)#.

In the given series #{540,90,15,....}#, first term is #540# and common ration #r# is #90/540=15/90=1/6#

Hence the sixth term #a_6# is #a_1xxr^(n-1)=540xx1/6^(6-1)#

= #540/(6xx6xx6xx6xx6)=(6xx6xx3xx5)/(6xx6xx6xx6xx6)#

= #(cancel6xxcancel6xxcancel3xx5)/(cancel6xxcancel6xx2cancel6xx6xx6)#

= #5/72#