What is the 9th term of the geometric sequence where a1 = 625 and a3 = 25?

1 Answer
Mar 9, 2016

Answer:

#a_9=1/625#

Explanation:

Let the geometric series be #{a_1,a_2,a_3,a_4,.....}#, where #(a_2)/(a_1)=(a_3)/(a_2)=(a_4)/(a_3)=(a_5)/(a_4)=.....=r#.

the #n^(th)# of such a series is given by #a_n=axxr^(n-1)#

As #a_1=625#, hence #a_3=625xxr^2#, but this is #25#.

Hence #625xxr^2=25# or #r^2=1/25# or #r=+-1/5#

Hence #9^(th)# term #a_9=625xx(+-1/5^8)#

or #a_9=625/25^4=1/625#