Question

Dada a PG (-240, -120, -60, ...), determine sua razão e o 6o termo.

Answer 1

$\boxed{\boxed{\boxed{\mathsf{q=\dfrac{-120}{-240}=\dfrac{1}{2}\checkmark}}}}$

$\mathsf{a_{6}=a_{3}.q^3}$

$\mathsf{a_{6}=-60\times(\dfrac{1}{2})^3}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{a_{6}= - \dfrac{15}{2}}}}}}$