Question

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

Answer 1

Answer:

$t_7 =\frac{5}{729}$

Step-by-step explanation:

We are given that:

First term (a) = 5

common ratio (r) =$\frac{1}{3}$

The nth term of a geometric sequence is given as:

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