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引例一个等差数列的前10项和为100,前100项和为10,求前110项之和.设等差数列为{a_n},前n项和为S_n,此题的求解方法多种多样:可以由S_(10)=100和S_(100)=10建立首项a_1与公差d的方程组,解得a_1与d,进而求出S_(110);可设S_n=An~2+Bn,由S_(10)=100和S_(100)=10解得S_n的通项,即得S_(110);也可利用S_(10),S_(20)-S_(10),…,S_(100)-S_(90),S_(110)-S_(100)成等差数列;或者利用
For example, the first 10 items of an arithmetic sequence are 100, the first 100 items and the last item of 10, and the sum of the first 110 items is set as {a_n}, and the first n items and the latter are S_n. There are many ways to solve this problem Diverse: The first equation of a_1 and the tolerance d can be established from S_ (10) = 100 and S_ (100) = 10 to solve for a_1 and d, then find S_ (110); we can set S_n = An ~ 2 + Bn, the general term of S_n is obtained by S_ (10) = 100 and S_ (100) = 10, that is S_ (110) S_ (100) -S_ (90), S_ (110) -S_ (100) into arithmetic progression; or