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            • 1.
              已知\(\{a_{n}\}\)是等差数列,\(\{b_{n}\}\)是各项为正的等比数列,且\(a_{1}=b_{1}=1\),\(a_{3}+b_{5}=21\),\(a_{5}+b_{3}=13\).
              \((1)\)求数列\(\{a_{n}\}\),\(\{b_{n}\}\)的通项公式;
              \((2)\)求数列\(\{a_{n}+b_{n}\}\) 的前\(n\)项和\(S_{n}\).
            • 2.
              在\(\triangle ABC\)中,角\(A\),\(B\),\(C\)的对边分别为\(a\),\(b\),\(c\),若\(A\),\(B\),\(C\)成等差数列,\(2a\),\(2b\),\(2c\)成等比数列,则\(\cos A\cos B=(\)  \()\)
              A.\( \dfrac {1}{4}\)
              B.\( \dfrac {1}{6}\)
              C.\( \dfrac {1}{2}\)
              D.\( \dfrac {2}{3}\)
            • 3.
              已知等差数列\(\{a_{n}\}\)的公差\(d\neq 0\),且\(a_{1}\)、\(a_{3}\)、\(a_{9}\)成等比数列,则\( \dfrac {a_{1}+a_{3}+a_{9}}{a_{2}+a_{4}+\;a_{10}}\)的值为\((\)  \()\)
              A.\( \dfrac {9}{14}\)
              B.\( \dfrac {11}{15}\)
              C.\( \dfrac {13}{16}\)
              D.\( \dfrac {15}{17}\)
            • 4.
              \(a_{1}\),\(a_{2}\),\(a_{3}\),\(a_{4}\)是各项不为零的等差数列且公差\(d\neq 0\),若将此数列删去某一项得到的数列\((\)按原来的顺序\()\)是等比数列,则\( \dfrac {a_{1}}{d}\)的值为\((\)  \()\)
              A.\(-4\)或\(1\)
              B.\(1\)
              C.\(4\)
              D.\(4\)或\(-1\)
            • 5.
              已知等比数列\(\{a_{n}\}\)中,各项都是正数,且\(a_{1}\),\( \dfrac {1}{2}a_{3}\),\(2a_{2}\)成等差数列,则\( \dfrac {a_{9}+a_{10}}{a_{7}+a_{8}}=(\)  \()\)
              A.\(1+ \sqrt {2}\)
              B.\(1- \sqrt {2}\)
              C.\(3+2 \sqrt {2}\)
              D.\(3-2 \sqrt {2}\)
            • 6.
              若等差数列\(\{a_{n}\}\)的公差为\(2\),且\(a_{5}\)是\(a_{2}\)与\(a_{6}\)的等比中项,则该数列的前\(n\)项和\(S_{n}\)取最小值时,\(n\)的值等于\((\)  \()\)
              A.\(4\)
              B.\(5\)
              C.\(6\)
              D.\(7\)
            • 7.
              已知下列三角形数表假设第行的第二个数为\(a_{n}(n\geqslant 2,n∈N^{*}).\)
              \((1)\)依次写出第六行的所有数字;
              \((2)\)归纳出\(a_{n+1}\)与\(a_{n}\)的关系式并求出\(a_{n}\)的通项公式;
              \((3)\)设\(a_{n}⋅b_{n}=1\),求证:\(b_{1}+b_{2}+b_{3}+…+b_{n} < 2\).
            • 8.
              如果一个数列从第\(2\)项起,每一项与它前一项的差都大于\(3\),则称这个数列为“\(S\)型数列”.
              \((1)\)已知数列\(\{a_{n}\}\)满足\(a_{1}=4\),\(a_{2}=8\),\(a_{n}+a_{n-1}=8n-4(n\geqslant 2,n∈N^{*})\),求证:数列\(\{a_{n}\}\)是“\(S\)型数列”;
              \((2)\)已知等比数列\(\{a_{n}\}\)的首项与公比\(q\)均为正整数,且\(\{a_{n}\}\)为“\(S\)型数列”,记\(b_{n}= \dfrac {3}{4}a_{n}\),当数列\(\{b_{n}\}\)不是“\(S\)型数列”时,求数列\(\{a_{n}\}\)的通项公式;
              \((3)\)是否存在一个正项数列\(\{c_{n}\}\)是“\(S\)型数列”,当\(c_{2}=9\),且对任意大于等于\(2\)的自然数\(n\)都满足\(( \dfrac {1}{n}- \dfrac {1}{n+1})(2+ \dfrac {1}{c_{n}})\leqslant \dfrac {1}{c_{n-1}}+ \dfrac {1}{c_{n}}\leqslant ( \dfrac {1}{n}- \dfrac {1}{n+1})(2+ \dfrac {1}{c_{n-1}})\)?如果存在,给出数列\(\{c_{n}\}\)的一个通项公式\((\)不必证明\()\);如果不存在,请说明理由.
            • 9.
              已知数列\(\{a_{n}\}\)是等比数列,\(a_{2}=4\),\(a_{3}+2\)是\(a_{2}\)和\(a_{4}\)的等差中项.
              \((\)Ⅰ\()\)求数列\(\{a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)设\(b_{n}=2\log _{2}a_{n}-1\),求数列\(\{a_{n}b_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 10.
              设\(\{a_{n}\}\)是公差不为零的等差数列,满足\(a_{6}=5\),\(a_{2}^{2}+a_{3}^{2}=a_{4}^{2}+a_{5}^{2}\),数列\(\{b_{n}\}\)的通项公式为\(b_{n}=3n-11\)
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)若从数列\(\{a_{n}\}\),\(\{b_{n+4}\}\)中按从小到大的顺序取出相同的项构成数列\(\{C_{n}\}\),直接写出数列\(\{C_{n}\}\)的通项公式;
              \((3)\)记\(d_{n}= \dfrac {b_{n}}{a_{n}}\),是否存在正整数\(m\),\(n(m\neq n\neq 5)\),使得\(d_{5}\),\(d_{m}\),\(d_{n}\)成等差数列?若存在,求出\(m\),\(n\)的值;若不存在,请说明理由.
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