Đáp án: $ q=\dfrac{11}{26}$
Giải thích các bước giải:
Ta có:
$u_1=2\to u_n=q^{n-1}u_1= \dfrac18\to q^{n-1}\cdot 2=\dfrac18\to q^{n-1}=\dfrac1{16}$
Lại có $S_{n}=\dfrac{27}{8}$
$\to u_1+u_2+...+u_n=\dfrac{27}{8}$
$\to u_1+qu_1+...+q^{n-1}u_1=\dfrac{27}{8}$
$\to u_1(1+q+...+q^{n-1})=\dfrac{27}{8}$
$\to u_1\cdot \dfrac{q^n-1}{q-1}=\dfrac{27}{8}$
$\to2\cdot \dfrac{q^n-1}{q-1}=\dfrac{27}{8}$
$\to \dfrac{q^n-1}{q-1}=\dfrac{27}{16}$
Lại có:
$q^{n-1}=\dfrac1{16}\to q^n=\dfrac1{16}q$
$\to \dfrac{\dfrac1{16}q-1}{q-1}=\dfrac{27}{16}$
$\to q=\dfrac{11}{26}$
