Question

Fond the 7th term of the geometric sequence whose common ratio is 1/3 and whose first term is 4

Answer 1

`a_7=(4)/(729)`

Explanation:

General form of geometric progression:
`a_n=ar^(n-1)`

(where
`a` is the initial term and
`r` is the common ratio)

Given:

• 
  `a=4`
• 
  `r=\frac13`

`\implies a_n=4\left(\frac13\right)^(n-1)`

Therefore, when n = 7:

`\implies a_7=4\left(\frac13\right)^(7-1)`

`\implies a_7=4\left(\frac13\right)^(6)`

`\implies a_7=4\left((1^6)/(3^6)\right)`

`\implies a_7=4\left((1)/(729)\right)`

`\implies a_7=(4)/(729)`