25 câu trắc nghiệm Toán 11 Kết nối tri thức bài 7: Cấp số nhân có đáp án

Xét đáp án A ta có: \[128; - 64;32; - 16; - 8;... \to \frac{{{{\rm{u}}_{\rm{2}}}}}{{{{\rm{u}}_{\rm{1}}}}}{\rm{ = }}\frac{{ - {\rm{1}}}}{{\rm{2}}}{\rm{ = }}\frac{{{{\rm{u}}_{\rm{3}}}}}{{{{\rm{u}}_{\rm{2}}}}}{\rm{ = }}\frac{{{{\rm{u}}_{\rm{4}}}}}{{{{\rm{u}}_{\rm{3}}}}}\].

Xét đáp án B ta có: \[\sqrt {\rm{2}} {\rm{; 2; 4; 4}}\sqrt {\rm{2}} {\rm{; }}... \to \frac{{{{\rm{u}}_{\rm{2}}}}}{{{{\rm{u}}_{\rm{1}}}}}{\rm{ = }}\sqrt {\rm{2}} \ne {\rm{2 = }}\frac{{{{\rm{u}}_{\rm{3}}}}}{{{{\rm{u}}_{\rm{2}}}}} \to \] loại B.

Xét đáp án C ta có: \[{\rm{5; 6; 7; 8;}}... \to \frac{{{{\rm{u}}_{\rm{2}}}}}{{{{\rm{u}}_{\rm{1}}}}}{\rm{ = }}\frac{{\rm{6}}}{{\rm{5}}} \ne \frac{{\rm{7}}}{{\rm{6}}}{\rm{ = }}\frac{{{{\rm{u}}_{\rm{3}}}}}{{{{\rm{u}}_{\rm{2}}}}} \to \] loại C.

Xét đáp án D ta có: \[{\rm{15; 5; 1; }}\frac{{\rm{1}}}{{\rm{5}}}{\rm{;}}... \to \frac{{{{\rm{u}}_{\rm{2}}}}}{{{{\rm{u}}_{\rm{1}}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}} \ne \frac{{\rm{1}}}{{\rm{5}}}{\rm{ = }}\frac{{{{\rm{u}}_{\rm{3}}}}}{{{{\rm{u}}_{\rm{2}}}}} \to \] loại D.

Ta có u₅ = u₁.q⁴ = \(2.{\left( {\frac{1}{2}} \right)^4} = \frac{1}{8}\).

Ta có u_n = u₁.q^{n – 1} = 5.2^{n – 1}.

Ta có u₁ = −2; u₂ = (−2).(−5) = 10; u₃ = 10.(−5) = −50; u₄ = (−50). (−5) = 250.

\[\frac{{ - 9}}{{\rm{x}}}{\rm{ = }}\frac{{\rm{x}}}{{ - 4}} \Leftrightarrow {{\rm{x}}^{\rm{2}}}{\rm{ = 36}} \Leftrightarrow {\rm{x = }} \pm 6\].

Ta có \({q^3} = \frac{{{u_4}}}{{{u_1}}} = \frac{{54}}{2} = 27 \Rightarrow q = 3\).

Có u_n = u₁.q^{n – 1}  Û 1.2^{n – 1} = 1024 Û 1.2^{n – 1} = 2¹⁰ Û n – 1 = 10 Û n = 11.

\(\left\{ {\begin{array}{*{20}{c}}{{\rm{192 = }}{{\rm{u}}_{\rm{6}}}{\rm{ = }}{{\rm{u}}_{\rm{1}}}{\rm{.}}{{\rm{q}}^{\rm{5}}}}\\{{\rm{384 = }}{{\rm{u}}_{\rm{7}}}{\rm{ = }}{{\rm{u}}_{\rm{1}}}{\rm{.}}{{\rm{q}}^{\rm{6}}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{{{{\rm{u}}_{\rm{7}}}}}{{{{\rm{u}}_{\rm{6}}}}}{\rm{ = }}\frac{{{{\rm{u}}_{\rm{1}}}{\rm{.}}{{\rm{q}}^{\rm{6}}}}}{{{{\rm{u}}_{\rm{1}}}{\rm{.}}{{\rm{q}}^{\rm{5}}}}}{\rm{ = }}\frac{{{\rm{384}}}}{{{\rm{192}}}}{\rm{ = 2}}}\\{{\rm{384 = }}{{\rm{u}}_{\rm{1}}}{\rm{.}}{{\rm{q}}^{\rm{6}}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{{\rm{u}}_{\rm{1}}}{\rm{ = 6}}}\\{{\rm{q = 2}}}\end{array}} \right.\).