Question

4．兩等差數(shù)列{a_n}和{b_n}，前n項(xiàng)和分別為S_n，T_n，且$\frac{{S}_{n}}{{T}_{n}}$=$\frac{7n+2}{n+3}$，則$\frac{{{a_2}+{a_{20}}}}{{{b_7}+{b_{15}}}}$等于$\frac{149}{24}$．

Answer 1

分析 利用等差數(shù)列的性質(zhì)可得$\frac{{{a_2}+{a_{20}}}}{{{b_7}+{b_{15}}}}$=$\frac{{a}_{1}+{a}_{21}}{_{1}+_{21}}$，進(jìn)而可得$\frac{{{a_2}+{a_{20}}}}{{{b_7}+{b_{15}}}}$=$\frac{{S}_{21}}{{T}_{21}}$，代入計(jì)算即得結(jié)論．

解答 解：∵數(shù)列{a_n}和{b_n}均為等差數(shù)列，
∴$\frac{{{a_2}+{a_{20}}}}{{{b_7}+{b_{15}}}}$=$\frac{{a}_{1}+{a}_{21}}{_{1}+_{21}}$=$\frac{\frac{21}{2}（{a}_{1}+{a}_{21}）}{\frac{21}{2}（_{1}+_{21}）}$=$\frac{{S}_{21}}{{T}_{21}}$，
又∵$\frac{{S}_{n}}{{T}_{n}}$=$\frac{7n+2}{n+3}$，
∴$\frac{{S}_{21}}{{T}_{21}}$=$\frac{7×21+2}{21+3}$=$\frac{149}{24}$，
∴$\frac{{{a_2}+{a_{20}}}}{{{b_7}+{b_{15}}}}$=$\frac{149}{24}$，
故答案為：$\frac{149}{24}$．

點(diǎn)評 本題考查等差數(shù)列的簡單性質(zhì)，注意解題方法的積累，屬于中檔題．