Đáp án:
Giải thích các bước giải:
$\bullet \,\,\,\,\,$${{u}_{1}}+{{u}_{6}}=244$
$\to {{u}_{1}}+{{u}_{1}}.{{q}^{5}}=244$
$\to {{u}_{1}}\left( 1+{{q}^{5}} \right)=244$
$\bullet \,\,\,\,\,$${{u}_{3}}+{{u}_{4}}=36$
$\to {{u}_{1}}.{{q}^{2}}+{{u}_{1}}.{{q}^{3}}=36$
$\to {{u}_{1}}\left( {{q}^{2}}+{{q}^{3}} \right)=36\,\,\,\,\,\left( 2 \right)$
Lấy $\frac{\left( 1 \right)}{\left( 2 \right)}$, ta được
$\to \frac{1+{{q}^{5}}}{{{q}^{2}}+{{q}^{3}}}=\frac{61}{9}$
$\to 9\left( 1+{{q}^{5}} \right)=61\left( {{q}^{2}}+{{q}^{3}} \right)$
$\to 9{{q}^{5}}-61{{q}^{3}}-61{{q}^{2}}+9=0$
$\to 9{{q}^{5}}-27{{q}^{4}}+27{{q}^{4}}-81{{q}^{3}}+20{{q}^{3}}-60{{q}^{2}}-{{q}^{2}}+3q-3q+9=0$
$\to 9{{q}^{4}}\left( q-3 \right)+27{{q}^{3}}\left( q-3 \right)+20{{q}^{2}}\left( q-3 \right)-q\left( q-3 \right)-3\left( q-3 \right)=0$
$\to \left( q-3 \right)\left( 9{{q}^{4}}+27{{q}^{3}}+20{{q}^{2}}-q-3 \right)$
$\to \left( q-3 \right)\left( 3{{q}^{2}}+2q-1 \right)\left( 3{{q}^{2}}+7q+3 \right)=0$
$\to q=3$ hoặc $q=\frac{1}{3}$ hoặc $q=-1$ hoặc $q=\frac{-7+\sqrt{13}}{6}$ hoặc $q=\frac{-7-\sqrt{13}}{6}$
$\bullet \,\,\,\,\,$Thế $q$ lần lượt vào công thức $\left( 1 \right)$ hoặc $\left( 2 \right)$, ta tìm được ${{u}_{1}}$
$\bullet \,\,\,\,\,$$q=3\to {{u}_{1}}=1$
$\bullet \,\,\,\,\,$$q=\frac{1}{3}\to {{u}_{1}}=243$
$\bullet \,\,\,\,\,$$q=-1\to {{u}_{1}}\in \varnothing $
$\bullet \,\,\,\,\,$$q=\frac{-7+\sqrt{13}}{6}\to {{u}_{1}}=38\sqrt{13}+122$
$\bullet \,\,\,\,\,$$q=\frac{-7-\sqrt{13}}{6}\to {{u}_{1}}=-38\sqrt{13}+122$
