Question

Find the 9th term in a geometric sequence whose common ration is 1/3 and first term is 6

Answer 1

Therefore 9th term of the geometric sequence is
`(2)/(2187)`

Explanation:

Given first term(a) = 6

and common ratio =
`(1)/(3)`

`T_n= ar^(n-1)`

`\therefore T_9=ar^(9-1)`

`\Leftrightarrow T_9=a r^8`

`\Leftrightarrow T_9=6* ( (1)/(3)) ^8`

`\Leftrightarrow T_9=(2)/(2187)`

Therefore 9th term of the geometric sequence is
`(2)/(2187)`