How do you find #a_9# when #a_1 = 1000, r = 1/3?#
1 Answer
May 25, 2016
Explanation:
For the standard geometric sequence.
#a,ar,ar^2,ar^3,..............,ar^(n-1)#
where a is 1st term
r ,the common ratio
#=a_2/a_1=a_3/a_2=....=a_n/a_(n-1)# and the nth term
#a_n=ar^(n-1)# here n = 9 , a = 1000 and
#r=1/3#
#rArra_9=1000(1/3)^8=1000/3^8=1000/6561#
