Question

Find the 8th term of the geometric sequence, given the first term and common ratio. a1=6 r=-1/3

Answer 1

The rule of a geometric sequence:

$a_n=a_1r^{n-1}$

We have:

$a_1=6;\ r=-\dfrac{1}{3}$

Find $a_8=?\to n=8$

substitute:

$a_8=6\cdot\left(-\dfrac{1}{3}\right)^{8-1}=6\cdot\left(-\dfrac{1}{3}\right)^7=6\cdot\left(-\dfrac{1}{2187}\right)=-\dfrac{2}{729}$