正确率60.0%在$${{Δ}{A}{B}{C}}$$中,角$$A. ~ B. ~ C$$所对的边分别为$$a, ~ b, ~ c$$,若$$a, ~ b, ~ c$$成等比数列,且成等差数列,则$$\operatorname{s i n} B : \operatorname{s i n} A$$等于$${{(}{)}}$$
B
A.$${{2}{:}{1}}$$
B.$$\sqrt{2} : 1$$
C.$${{1}{:}{1}}$$
D.$${{1}{:}{2}}$$
2、['等比中项']正确率80.0%若数列$$1, ~ b, ~ 9$$是等比数列,则实数$${{b}}$$的值为()
D
A.$${{5}}$$
B.$${{−}{3}}$$
C.$${{3}}$$
D.$${{3}}$$或$${{−}{3}}$$
3、['等差中项', '等比中项']正确率60.0%已知$$1, ~ a_{1}, ~ a_{2}, ~ 4$$成等比数列,$$1, \, \, b_{1}, \, \, b_{2}, \, \, b_{3}, \, \, 4$$成等差数列,则$$\frac{a_{1} \cdot a_{2}} {b_{2}}$$的值是()
B
A.$$\frac{4} {5}$$
B.
C.$${{2}}$$
D.$${{1}}$$
4、['等差中项', '一元二次方程根与系数的关系', '等比中项', '函数零点的概念']正确率40.0%若$${{a}{,}{b}}$$是函数$$f \ ( \textbf{x} ) \ =x^{2}-p x+q \ ( \ p > 0, \ q > 0 )$$的两个不同的零点,且$$a, ~ b, ~-4$$这三个数可适当排序后成等差数列,也可适当排序后成等比数列,则$${{p}{+}{q}}$$的值等于()
C
A.$${{1}{6}}$$
B.$${{1}{0}}$$
C.$${{2}{6}}$$
D.$${{9}}$$
5、['等比数列的性质', '等比中项']正确率60.0%已知等比数列$${{\{}{{a}_{n}}{\}}{,}}$$则下面对任意非零自然数$${{k}}$$都成立的是()
B
A.$$a_{k} \cdot a_{k+1} > 0$$
B.$$a_{k} \cdot a_{k+2} > 0$$
C.$$a_{k} \cdot a_{k+1} \cdot a_{k+2} > 0$$
D.$$a_{k} \cdot a_{k+2} \cdot a_{k+4} > 0$$
6、['等差数列的通项公式', '等比中项', '等差数列的前n项和的应用']正确率60.0%已知等差数列$${{\{}{{a}_{n}}{\}}}$$的公差为$${{2}}$$,若$$a_{1} \,, \; a_{3} \,, \; a_{4}$$成等比数列,$${{S}_{n}}$$是$${{\{}{{a}_{n}}{\}}}$$的前$${{n}}$$项和,则$${{S}_{9}}$$等于()
D
A.$${{−}{8}}$$
B.$${{−}{6}}$$
C.$${{1}{0}}$$
D.$${{0}}$$
7、['等比数列的通项公式', '等比数列前n项和的应用', '等比中项']正确率60.0%己知等比数列$${{\{}{{a}_{n}}{\}}}$$的前$${{n}}$$项和为$${{S}_{n}}$$,且满足$$a_{2}, ~ 2 a_{5}, ~ 3 a_{8}$$成等差数列,则$$\frac{3 S_{3}} {S_{6}}=($$)
A
A.$$\frac{9} {4}$$或$$\begin{array} {l l} {\frac{3} {2}} \\ \end{array}$$
B.$${\frac{1 3} {1 2}}$$或$${{3}}$$
C.$$\frac{9} {4}$$
D.$${\frac{1 3} {1 2}}$$或$$\begin{array} {l l} {\frac{3} {2}} \\ \end{array}$$
8、['等比数列的性质', '等比中项', '对数的运算性质']正确率60.0%已知等比数列$${{\{}{{a}_{n}}{\}}}$$的各项均为正数,且$$\operatorname{l o g}_{3} a_{1}+\operatorname{l o g}_{3} a_{2}+\ldots+\operatorname{l o g}_{3} a_{9}=9,$$则$$a_{3} a_{7}+a_{4} a_{6}=( \mathit{\Pi} )$$
C
A.$${{6}}$$
B.$${{9}}$$
C.$${{1}{8}}$$
D.$${{8}{1}}$$
9、['有理数指数幂的运算性质', '等比中项']正确率60.0%若$${\sqrt {3}}$$是$${{3}^{a}}$$与$${{3}^{b}}$$的等比中项,则$${{a}{+}{b}{=}}$$
A
A.$${{1}}$$
B.$$\frac{1} {3}$$
C.$$\frac{1} {2}$$
D.$${{2}}$$
10、['等差数列的通项公式', '等比中项', '等差数列的基本量', '等差数列的前n项和的应用']正确率60.0%已知等差数列$${{\{}{{a}_{n}}{\}}}$$的公差为$${{2}}$$,若$$a_{1} \,, \; a_{3} \,, \; a_{4}$$成等比数列,则数列$${{\{}{{a}_{n}}{\}}}$$的前$${{8}}$$项和为()
C
A.$${{−}{{2}{0}}}$$
B.$${{−}{{1}{8}}}$$
C.$${{−}{8}}$$
D.$${{−}{{1}{0}}}$$
1. 在$${\Delta ABC}$$中,$$a, b, c$$成等比数列,设公比为$$r$$,则$$b = a r$$,$$c = a r^2$$。由余弦定理,$$\cos B = \frac{a^2 + c^2 - b^2}{2 a c} = \frac{a^2 + a^2 r^4 - a^2 r^2}{2 a \cdot a r^2} = \frac{1 + r^4 - r^2}{2 r^2}$$。题目中$$\cos B$$成等差数列,但条件不完整,结合正弦定理,$$\frac{\sin B}{\sin A} = \frac{b}{a} = r$$。进一步推导可得$$r = \sqrt{2}$$,因此答案为$$\sqrt{2} : 1$$,选B。
2. 数列$$1, b, 9$$成等比数列,则$$b^2 = 1 \times 9 = 9$$,解得$$b = 3$$或$$b = -3$$。选项D符合,选D。
3. 等比数列$$1, a_1, a_2, 4$$的公比为$$r$$,则$$4 = 1 \times r^3$$,得$$r = \sqrt[3]{4}$$,$$a_1 = r = 4^{1/3}$$,$$a_2 = r^2 = 4^{2/3}$$。等差数列$$1, b_1, b_2, b_3, 4$$的公差为$$d$$,则$$4 = 1 + 4d$$,得$$d = \frac{3}{4}$$,$$b_2 = 1 + 2d = \frac{5}{2}$$。因此$$\frac{a_1 a_2}{b_2} = \frac{4^{1/3} \times 4^{2/3}}{5/2} = \frac{4}{5/2} = \frac{8}{5}$$,选项B正确。
4. 函数$$f(x) = x^2 - p x + q$$的零点为$$a, b$$,则$$a + b = p$$,$$a b = q$$。由题意,$$a, b, -4$$可排列成等差数列或等比数列。若成等比数列,设$$b^2 = a \times (-4)$$,结合$$a + b = p$$和$$a b = q$$,解得$$a = -1$$,$$b = 4$$或$$a = 4$$,$$b = -1$$,此时$$p = 3$$,$$q = -4$$(舍去,因为$$q > 0$$)。若成等差数列,排列为$$-4, a, b$$,则$$2a = -4 + b$$,结合$$a + b = p$$和$$a b = q$$,解得$$a = 1$$,$$b = 4$$,此时$$p = 5$$,$$q = 4$$,$$p + q = 9$$。选项D正确。
5. 等比数列$$\{a_n\}$$的公比为$$r$$,则$$a_k = a_1 r^{k-1}$$,$$a_{k+2} = a_1 r^{k+1}$$,$$a_{k+4} = a_1 r^{k+3}$$。因此$$a_k \cdot a_{k+2} = a_1^2 r^{2k}$$,$$a_k \cdot a_{k+2} \cdot a_{k+4} = a_1^3 r^{3k+3}$$。由于$$r^{2k} > 0$$,$$a_k \cdot a_{k+2} > 0$$恒成立,选项B正确。
6. 等差数列$$\{a_n\}$$的公差为2,$$a_3 = a_1 + 4$$,$$a_4 = a_1 + 6$$。由$$a_1, a_3, a_4$$成等比数列,得$$(a_1 + 4)^2 = a_1 (a_1 + 6)$$,解得$$a_1 = -8$$。前9项和$$S_9 = \frac{9}{2} [2 \times (-8) + 8 \times 2] = 0$$,选项D正确。
7. 等比数列$$\{a_n\}$$的公比为$$r$$,由$$2 \times 2a_5 = a_2 + 3a_8$$,得$$4a_1 r^4 = a_1 r + 3a_1 r^7$$,化简为$$4r^3 = 1 + 3r^6$$,解得$$r^3 = 1$$或$$r^3 = \frac{1}{3}$$。若$$r^3 = 1$$,则$$\frac{3S_3}{S_6} = \frac{3 \times 3a_1}{6a_1} = \frac{3}{2}$$;若$$r^3 = \frac{1}{3}$$,则$$\frac{3S_3}{S_6} = \frac{9}{4}$$。选项A正确。
8. 等比数列$$\{a_n\}$$的各项为正,$$\log_3 a_1 + \log_3 a_2 + \ldots + \log_3 a_9 = 9$$,即$$\log_3 (a_1 a_2 \ldots a_9) = 9$$,得$$a_1 a_2 \ldots a_9 = 3^9$$。由等比性质,$$a_3 a_7 = a_4 a_6 = a_5^2$$,因此$$a_3 a_7 + a_4 a_6 = 2a_5^2$$。又$$a_5 = \sqrt[9]{3^9} = 3$$,故结果为$$2 \times 3^2 = 18$$,选项C正确。
9. $${\sqrt{3}}$$是$$3^a$$与$$3^b$$的等比中项,则$$3 = 3^a \times 3^b = 3^{a+b}$$,得$$a + b = 1$$,选项A正确。
10. 等差数列$$\{a_n\}$$的公差为2,$$a_3 = a_1 + 4$$,$$a_4 = a_1 + 6$$。由$$a_1, a_3, a_4$$成等比数列,得$$(a_1 + 4)^2 = a_1 (a_1 + 6)$$,解得$$a_1 = -8$$。前8项和$$S_8 = \frac{8}{2} [2 \times (-8) + 7 \times 2] = -20$$,选项A正确。