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![S_n=\dfrac{n}{2}[2a+(n-1)d] \\$Therefore$:\\\\S_{40}=\dfrac{40}{2}[2(4)+(40-1)*8] \\=20(8+39*8)\\=20(8+312)\\=20*320\\=6560](/tpl/images/0702/3597/86c68.png)
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Given 4,12,20,28 find s40...
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