已知等差数列{an}的公差为-2,若a3+a6+a9+...+a99=-82,则a1+a4+a7+...+a97等于___.帮忙分析一下解析【解:a3+a6+a9+...+a99 =a1+2d+a4+2d+...+a97+2d =(a1+a4+...+a97)+33*2d=-82 所以a1+a4+a7+...+a97 =-82+33*4
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![已知等差数列{an}的公差为-2,若a3+a6+a9+...+a99=-82,则a1+a4+a7+...+a97等于___.帮忙分析一下解析【解:a3+a6+a9+...+a99 =a1+2d+a4+2d+...+a97+2d =(a1+a4+...+a97)+33*2d=-82 所以a1+a4+a7+...+a97 =-82+33*4](/uploads/image/z/5444947-19-7.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E5%B7%AE%E4%B8%BA-2%2C%E8%8B%A5a3%2Ba6%2Ba9%2B...%2Ba99%3D-82%2C%E5%88%99a1%2Ba4%2Ba7%2B...%2Ba97%E7%AD%89%E4%BA%8E___.%E5%B8%AE%E5%BF%99%E5%88%86%E6%9E%90%E4%B8%80%E4%B8%8B%E8%A7%A3%E6%9E%90%E3%80%90%E8%A7%A3%EF%BC%9Aa3%2Ba6%2Ba9%2B...%2Ba99++++++++%3Da1%2B2d%2Ba4%2B2d%2B...%2Ba97%2B2d++++++++%3D%28a1%2Ba4%2B...%2Ba97%29%2B33%2A2d%3D-82+++%E6%89%80%E4%BB%A5a1%2Ba4%2Ba7%2B...%2Ba97++++++++%3D-82%2B33%2A4)
已知等差数列{an}的公差为-2,若a3+a6+a9+...+a99=-82,则a1+a4+a7+...+a97等于___.帮忙分析一下解析【解:a3+a6+a9+...+a99 =a1+2d+a4+2d+...+a97+2d =(a1+a4+...+a97)+33*2d=-82 所以a1+a4+a7+...+a97 =-82+33*4
已知等差数列{an}的公差为-2,若a3+a6+a9+...+a99=-82,则a1+a4+a7+...+a97等于___.帮忙分析一下解析
【解:a3+a6+a9+...+a99
=a1+2d+a4+2d+...+a97+2d
=(a1+a4+...+a97)+33*2d=-82
所以a1+a4+a7+...+a97
=-82+33*4
=50】
【问:怎么知道是33个2d的?】
已知等差数列{an}的公差为-2,若a3+a6+a9+...+a99=-82,则a1+a4+a7+...+a97等于___.帮忙分析一下解析【解:a3+a6+a9+...+a99 =a1+2d+a4+2d+...+a97+2d =(a1+a4+...+a97)+33*2d=-82 所以a1+a4+a7+...+a97 =-82+33*4
因为一共有33个项啊.如果一定要计算
你可以将a3 a6 a9 ```a99 看成一个首项a1为3,公差d为3的等差数列
an=a1+(n-1)d
99=3+3(n-1)
99=3n
n=33
便可知道99是这个数列的第33项
所以你上面的题里有33个2d