一、等差数列的性质
待后补充,等等,
补充:⑦若数列\(\{a_n\}\)为等差数列,且公差\(d\neq 0\),则数列\(\{\cfrac{S_n}{n}\}\)也为等差数列;
分析:\(S_n=An^2+Bn(A\neq 0)\),则\(\cfrac{S_n}{n}=An+B\),则数列\(\{\cfrac{S_n}{n}\}\)也为等差数列;
⑧两个等差数列\(\{a_n\}\)和\(\{b_n\}\)的前\(n\)项和分别为\(S_n\)和\(T_n\),则有\(\cfrac{S_{2n-1}}{T_{2n-1}}=\cfrac{a_n}{b_n}\);
证明:由于等差数列\(\{a_n\}\)和\(\{b_n\}\)的前\(n\)项和分别为\(S_n\)和\(T_n\),
则\(S_{2n-1}=(2n-1)a_n\),\(T_{2n-1}=(2n-1)b_n\),故\(\cfrac{S_{2n-1}}{T_{2n-1}}=\cfrac{a_n}{b_n}\);
二、典例剖析:
\(\fbox{例1}\)【2018福建龙岩市高三质检】
已知等差数列\(\{a_n\}\)和\(\{b_n\}\),满足\(a_1+b_{10}=9\),\(a_3+b_8=15\),则\(a_5+b_6\)=______________.
分析:由已知得到,\(a_3+b_8=\cfrac{2a_3+2b_8}{2}\)
\(=\cfrac{(a_1+a_5)+(b_{10}+b_6)}{2}=\cfrac{(a_1+b_{10})+(a_5+b_6)}{2}\)
即\(15=\cfrac{9+(a_5+b_6)}{2}\),解得\(a_5+b_6=21\)。