优优班--学霸训练营 > 知识点挑题
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            • 1.
              已知等差数列\(\{a_{n}\}\)的公差为正数,且\(a_{3}⋅a_{7}=-12\),\(a_{4}+a_{6}=-4\),则\(S_{20}\)为\((\)  \()\)
              A.\(90\)
              B.\(-180\)
              C.\(180\)
              D.\(-90\)
            • 2.
              在等差数列\(\{a_{n}\}\)中,\(a_{9}= \dfrac {1}{2}a_{12}+3\),则数列\(\{a_{n}\}\)的前\(11\)项和\(S_{11}=(\)  \()\)
              A.\(24\)
              B.\(48\)
              C.\(66\)
              D.\(132\)
            • 3.
              已知\(\{a_{n}\}\)为等差数列,\(a_{1}+a_{3}+a_{5}=105\),\(a_{2}+a_{4}+a_{6}=99\),以\(S_{n}\)表示\(\{a_{n}\}\)的前\(n\)项和,则使得\(S_{n}\)达到最大值的\(n\)是\((\)  \()\)
              A.\(21\)
              B.\(20\)
              C.\(19\)
              D.\(18\)
            • 4.
              已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),\(a_{3}=7\),\(S_{9}=27\).
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)若\(b_{n}=|a_{n}|\),求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 5.
              在等差数列\(\{a_{n}\}\)中,\(a_{3}+a_{4}=15\),\(a_{2}a_{5}=54\),公差\(d < 0\).
              \((1)\)求数列\(\{a_{n}\}\)的通项公式\(a_{n}\);
              \((2)\)求数列的前\(n\)项和\(S_{n}\)的最大值及相应的\(n\)值.
            • 6.
              等差数列\(\{a_{n}\}\)中,\(a_{2}= \dfrac {5}{3},a_{6}=- \dfrac {7}{3}\),则\(\{a_{n}\}\)的前\(7\)项和\(S_{7}=\) ______ .
            • 7.
              已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),等比数列\(\{b_{n}\}\)的前\(n\)项和为\(T_{n}\),\(a_{1}=-1\),\(b_{1}=1\),\(a_{2}+b_{2}=2\).
              \((1)\)若\(a_{3}+b_{3}=5\),求\(\{b_{n}\}\)的通项公式;
              \((2)\)若\(T_{3}=21\),求\(S_{3}\).
            • 8.
              设\(S_{n}\)是等差数列\(\{a_{n}\}\)的前\(n\)项和,若\(a_{2}+a_{12}=18\),则\(S_{13}=(\)  \()\)
              A.\(91\)
              B.\(126\)
              C.\(234\)
              D.\(117\)
            • 9.
              已知等差数列\(\{a_{n}\}\)中,\(a_{3}=13\),\(a_{6}=25\)
              \((\)Ⅰ\()\)求\(\{a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)令\(b_{n}=2\;^{a_{n}}\),求证数列\(\{b_{n}\}\)是等比数列,并求\(\{b_{n}\}\)的前\(n\)项和\(S_{n}\).
            • 10.
              已知等差数列\(\{a_{n}\}\)满足:\(a_{3}=7\),\(a_{5}+a_{7}=26\),\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\).
              \((\)Ⅰ\()\)求\(a_{n}\)及\(S_{n}\);
              \((\)Ⅱ\()\)令\(b_{n}= \dfrac {1}{a_{n}^{2}-1}(n∈N^{*})\),求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\).
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