4 years ago

Helppppppppppppppppppp

1 answer:

Crank4 years ago

6 0

$\boxed{a_1=0.2}\\\\\\r=\dfrac{-0.06}{0.2}=\boxed{-0.3}$

so:

$a_n=a_1(r)^{(n-1)}\\\\\boxed{a_n=0.2(-0.3)^{(n-1)}}$

Answer E.

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