Question

A geometric sequence is shown below.

Answer 1

$a_1=2;\ a_2=-6;\ a_3=18;\ a_4=-54;\ a_5=162;\ ...$

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A recursive rule for a geometric sequence:

$a_1\\\\a_n=r\cdot a_{n-1}$

$r=\dfrac{a_{n+1}}{a_n}\to r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...\\\\r=\dfrac{-6}{2}=-3$

Therefore $\boxed{a_1=2;\qquad a_n=-3a_{n-1}}$

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The exciplit rule:

$a_n=a_1r^{n-1}$

Substitute:

$a_n=2(-3)^{n-1}=2(3)^n(3)^{-1}=2(3)^n\left(\dfrac{1}{3}\right)\\\\\boxed{a_n=\dfrac{2}{3}\left(-3)^n}$