Đáp án:
Giải thích các bước giải:
$\bullet \,\,{{u}_{1}}+{{u}_{2}}+{{u}_{3}}=9$
$\to {{u}_{1}}+{{u}_{1}}+d+{{u}_{1}}+2d=9$
$\to 3{{u}_{1}}+3d=9$
$\to {{u}_{1}}+d=3$
$\to {{u}_{1}}=3-d$
$\bullet \,\,{{u}_{3}}^{2}+{{u}_{5}}^{2}=4$
$\to {{\left( {{u}_{1}}+2d \right)}^{2}}+{{\left( {{u}_{1}}+4d \right)}^{2}}=4$
$\to {{\left( 3-d+2d \right)}^{2}}+{{\left( 3-d+4d \right)}^{2}}=4$
$\to {{\left( 3+d \right)}^{2}}+{{\left( 3+3d \right)}^{2}}=4$
$\to {{d}^{2}}+6d+9+9{{d}^{2}}+18d+9=4$
$\to 10{{d}^{2}}+24d-10=0$
$\to d=-1$ hoặc $d=-\frac{7}{5}$
Với $d=-1\to {{u}_{1}}=4$$\to {{u}_{10}}={{u}_{1}}+9d=4+9\left( -1 \right)=-5$
Tổng ${{S}_{10}}=\frac{10\left( {{u}_{1}}+{{u}_{n}} \right)}{2}=\frac{10\left( 4-5 \right)}{2}=-5$
Với $d=-\frac{7}{5}\to {{u}_{1}}=\frac{22}{5}$$\to {{u}_{10}}={{u}_{1}}+9d=\frac{22}{5}+9.\left( -\frac{7}{5} \right)=-\frac{41}{5}$
Tổng ${{S}_{10}}=\frac{10\left( {{u}_{1}}+{{u}_{10}} \right)}{2}=\frac{10\left( \frac{22}{5}-\frac{41}{5} \right)}{2}=-19$
