Question

No PG(3,14,48), encontre

b)A soma dos 7 primeiro termos​

Answer 1

$\left\{a_1,a_2,a_3,\dots\right\}=\left\{3,12,48,\dots\right\}$

$\boxed{a_n=a_kq^{n-k}}\ \to\ a_2=a_1q\ \to$

$q=\dfrac{a_2}{a_1}=\dfrac{12}{3}\ \therefore\ \boxed{q=4}$

$\boxed{S_n=a_1\dfrac{q^n-1}{q-1}}\ \to\ S_7=a_1\dfrac{q^7-1}{q-1}\ \to$

$S_7=3\dfrac{4^7-1}{4-1}=3\dfrac{16383}{3}\ \therefore\ \boxed{S_7=16383}$