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Find the 7th term of the geometric sequence whose common ratio is 1/3 and whose first term is 5.

User Tommasop
by
7.8k points

1 Answer

1 vote

Answer:


t_7 =(5)/(729)

Explanation:

We are given that:

First term (a) = 5

common ratio (r) =
(1)/(3)

The nth term of a geometric sequence is given as:


t_n =ar^(n-1) \\\therefore t_7 =5* \bigg((1)/(3)\bigg) ^(7-1) \\\therefore t_7 =5* \bigg((1)/(3)\bigg) ^(6) \\\therefore t_7 =5* \bigg((1)/(3)\bigg) ^(6) \\\therefore t_7 =5* (1)/(729)\\\\\huge\red{\boxed{\therefore t_7 =(5)/(729) }}

User Darin
by
8.3k points

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