Question

1．已知等差數(shù)列{a_n}和{b_n} 的前n項(xiàng)和S分別為S_n、T_n，且$\frac{{S}_{n}}{{T}_{n}}$=$\frac{7n+1}{n+3}$，則$\frac{{a}_{2}+{a}_{5}+{a}_{17}+{a}_{22}}{_{8}+_{10}+_{12}+_{16}}$=（　�。�

• A． $\frac{31}{5}$
• B． $\frac{32}{5}$
• C． 6
• D． 7

Answer 1

分析 由已知利用等差數(shù)列的通項(xiàng)公式先求出$\frac{{a}_{2}+{a}_{5}+{a}_{17}+{a}_{22}}{_{8}+_{10}+_{12}+_{16}}$=$\frac{{a}_{1}+{a}_{22}}{_{1}+_{22}}$，再由等差數(shù)列前n項(xiàng)和公式推導(dǎo)出原式等于$\frac{{S}_{22}}{{T}_{22}}$，由此能求解出結(jié)果．

解答 解：∵等差數(shù)列{a_n}和{b_n} 的前n項(xiàng)和S分別為S_n、T_n，且$\frac{{S}_{n}}{{T}_{n}}$=$\frac{7n+1}{n+3}$，
∴由等差數(shù)列的通項(xiàng)公式可得：
$\frac{{a}_{2}+{a}_{5}+{a}_{17}+{a}_{22}}{_{8}+_{10}+_{12}+_{16}}$
=$\frac{2（2{a}_{1}+21d）}{2（2_{1}+21d）}$
=$\frac{{a}_{1}+{a}_{22}}{_{1}+_{22}}$
=$\frac{\frac{22（{a}_{1}+{a}_{22}）}{2}}{\frac{22（_{1}+_{22}）}{2}}$
=$\frac{{S}_{22}}{{T}_{22}}$=$\frac{7×22+1}{22+3}$
=$\frac{155}{25}$=$\frac{31}{5}$，
故選：A．

點(diǎn)評 本題考查兩個(gè)等差數(shù)列的若干項(xiàng)的和的比值的求法，是中檔題，解題時(shí)要認(rèn)真審題，注意等差數(shù)列的性質(zhì)的合理運(yùn)用．