How do you find #a_9# for the geometric sequence #1/5, 1, 5,...#?

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Answer 1 · 2016-10-23T19:53:41

.#a_9=5^7=78125#

The GP #1/5, 1, 5....#

comparing with the standard GP # a, ar, ar^2, ....#

we note the first term

#=1/5#

and the common ratio is #r= 5#

#:. a_n=ar^n-1#

so #a_9=(1/5)5^8#

ie..#a_9=5^7=78125#