Question

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

Answer 1

Answer:

Therefore 9th term of the geometric sequence is $\frac{2}{2187}$

Step-by-step explanation:

Given first term(a) = 6

and common ratio = $\frac{1}{3}$

$T_n= ar^{n-1}$

$\therefore T_9=ar^{9-1}$

$\Leftrightarrow T_9=a r^8$

$\Leftrightarrow T_9=6\times ( \frac{1}{3}) ^8$

$\Leftrightarrow T_9=\frac{2}{2187}$

Therefore 9th term of the geometric sequence is $\frac{2}{2187}$