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

Question

asked Aug 4, 2021 60.1k views

1 Answer

Darshan Sawardekar · Answer 1 · 2021-08-08T14:05:03+0000

Answer:

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)