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            • 1.

              已知等比数列\(\{a_{n}\}\)的公比\(q > 1\),\(a_{1}=1\),且\(2a_{2}\),\(a_{4}\),\(3a_{3}\)成等差数列.

              \((1)\)求数列\(\{a_{n}\}\)的通项公式;

              \((2)\)记\(b_{n}=2na_{n}\),求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\).

            • 2.
              若等差数列\(\{a_{n}\}\)中有\(a_{2}+a_{4024}=4\),则\(a_{2013}=(\)  \()\)
              A.\(2\)
              B.\(4\)
              C.\(3\)
              D.\(6\)
            • 3. 等差数列\(\{ \)\(a_{n}\)\(\}\)中, \(a\)\({\,\!}_{1}+\) \(a\)\({\,\!}_{4}+\) \(a\)\({\,\!}_{7}=39\), \(a\)\({\,\!}_{2}+\) \(a\)\({\,\!}_{5}+\) \(a\)\({\,\!}_{8}=33\),则 \(a\)\({\,\!}_{3}+\) \(a\)\({\,\!}_{6}+\) \(a\)\({\,\!}_{9}\)的值为\((\)  \()\)
              A.\(30\)    
              B.\(27\)    
              C.\(24\)     
              D.\(21\)
            • 4.

              等差数列\(\left\{ {{a}_{n}} \right\}\)中,\({{a}_{3}}=4\),前\(11\)项的和\(S_{11}=110\)则\(a_{9}=\)

              A.\(10\)
              B.\(12\)
              C.\(14\)
              D.\(16\)
            • 5.

              设正项等比数列\(\{{a}_{n}\} \)的前\(n \)项和为\({{S}_{n}}\),若\(S_{3}=3\),\(S_{9}-S_{6}=12\),则\(S_{6}=\)       

            • 6.
              已知\(4a_{n+1}-4a_{n}-9=0\),则数列\(\{a_{n}\}\)是\((\)  \()\)
              A.公差为\(9\)的等差数列
              B.公差为\( \dfrac {9}{4}\)的等差数列
              C.公差为\(4\) 的等差数列
              D.不是等差数列
            • 7.

              设数列\(\left\{ {{a}_{n}} \right\}\)满足对任意的\(n\in {{N}^{*}}\),\({{P}_{n}}\left( n,{{a}_{n}} \right)\)满足\(\overrightarrow{{{P}_{n}}{{P}_{n+1}}}=(1,2)\),且\({{a}_{1}}+{{a}_{2}}=4\),则数列\(\left\{ \dfrac{1}{{{a}_{n}}\cdot {{a}_{n+1}}} \right\}\)的前\(n\)项的和\({{S}_{n}}\)为_________.

            • 8.

              黑白两种颜色的正六边形地面砖按如图的规律拼成若干个图案:则第\(n\)个图案中有白色地面砖              块\(.\)

            • 9.
              在\(\triangle ABC\)中,三内角\(A\)、\(B\)、\(C\)成等差数列,则角\(B\)等于\((\)  \()\)
              A.\(30^{\circ}\)
              B.\(60^{\circ}\)
              C.\(90^{\circ}\)
              D.\(120^{\circ}\)
            • 10.
              设\(Sn\)是等差数列\(\{a_{n}\}\)的前\(n\)项和,若\(a_{1}+a_{3}+a_{5}=3\) ,则\(S_{5} =\)      \((\)   \()\)
              A.\(5\)       
              B.\(7\)      
              C.\(9\)          
              D.\(11\)
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