Question

aliguem pode me ajudar nessa PG​

Answer 1

$(a_1,a_2,\dots,a_n)=\bigg(\dfrac{3}{2},\dfrac{3}{4},\dots,\dfrac{3}{128}\bigg)$

$\boxed{a_n=a_kq^{n-k}}$

$a_2=a_1q\ \therefore\ \dfrac{3}{4}=\dfrac{3}{2}q\ \therefore\ \boxed{q=\dfrac{1}{2}}$

$a_n=a_1q^{n-1}\ \therefore\ \dfrac{3}{128}=\dfrac{3}{2}\bigg(\dfrac{1}{2}\bigg)^{n-1}\ \therefore\ \dfrac{1}{64}=\bigg(\dfrac{1}{2}\bigg)^{n-1}\ \therefore$

$2^{n-1}=2^6\ \therefore\ n-1=6\ \therefore\ \boxed{n=7}$