Question

find the seventh term of the geometric sequence whose nth term is an=-4(2/3)^n-1

Answer 1

Answer:

Answer is on the above image.

Answer 2

Given,

nth term of geometric sequence, $\[{a_n} = 4 \times {\left( {\frac{2}{3}} \right)^{n - 1}}\]$

Solution,

Calculate the 7th term of geometric sequence.

$\[{a_n} = 4 \times {\left( {\frac{2}{3}} \right)^{n - 1}}\]$

Put $\[n = 7\]$,

$\[ \Rightarrow {a_7} = 4 \times {\left( {\frac{2}{3}} \right)^{7 - 1}}\]$

$\[ \Rightarrow {a_7} = 4 \times {\left( {\frac{2}{3}} \right)^6}\]$

$\[ \Rightarrow {a_7} = \frac{{256}}{{729}}\]$

Hence the 7th term of geometric sequence is $\[\frac{{256}}{{729}}\]$.