Question

calcule o décimo termo da pg onde a1=-2 e a2=-6

Answer 1

$q = \dfrac{-6}{-2} \\ \\ \\ q = 3 \\ \\ \\ a_{10} = a_1 \cdot q^9 \\ \\ a_{10} = -2\cdot3^9 \\ \\ a_{10} = -2\cdot19683 \\ \\ \boxed{a_{10} = 39.366}$

Answer 2

$q= \frac{a_2}{a_1} \\ \\ q= \frac{-6}{-2} \\ \\ q=3$


$a_{10}=a_1*q^9 \\ \\ a_{10}=-2*(3)^9 \\ \\ a_{10}=-2*19683 \\ \\ a_{10}=-39366$