题目内容
5.在公差不为0的等差数列{an}中,a1,a4,a8成等比数列.(1)已知数列{an}的前6项和为23,求数列{an}的通项公式;
(2)若${b_n}=\frac{1}{{{a_n}{a_{n+1}}}}$,且数列{bn}的前n项和为Tn,若${T_n}=\frac{1}{9}-\frac{1}{n+9}$,求数列{an}的公差.
分析 首先由题意求出等差数列的首项和公差的关系.
(1)由等差数列的前6项和为23列式求得首项和公差,则数列{an}的通项公式可求;
(2)把${b_n}=\frac{1}{{{a_n}{a_{n+1}}}}$裂项,求其前n项和,由${T_n}=\frac{1}{9}-\frac{1}{n+9}$即可求得公差d的值.
解答 解:设等差数列{an}的公差为d,由a1,a4,a8成等比数列可得,${{a}_{4}}^{8}={a}_{1}{a}_{8}$,
即$({a}_{1}+3d)^{2}={a}_{1}({a}_{1}+7d)$,
∴${{a}_{1}}^{2}+6{a}_{1}d+9{d}^{2}={{a}_{1}}^{2}+7{a}_{1}d$,而d≠0,∴a1=9d.
(1)由数列{an}的前6项和为23,得${S}_{6}=6{a}_{1}+\frac{6×5d}{2}=23$,
即6a1+15d=23,∴54d+15d=23,故d=$\frac{1}{3}$,a1=3,
故数列{an}的通项公式为${a}_{n}=3+\frac{1}{3}(n-1)$=$\frac{1}{3}(n+8)$;
(2)${b_n}=\frac{1}{{{a_n}{a_{n+1}}}}$=$\frac{1}{d}(\frac{1}{{a}_{n}}-\frac{1}{{a}_{n+1}})$,
则数列{bn}的前n项和为Tn=$\frac{1}{d}[(\frac{1}{{a}_{1}}-\frac{1}{{a}_{2}})+(\frac{1}{{a}_{2}}-\frac{1}{{a}_{3}})+…+(\frac{1}{{a}_{n}}-\frac{1}{{a}_{n+1}})]$
=$\frac{1}{d}(\frac{1}{{a}_{1}}-\frac{1}{{a}_{n+1}})=\frac{1}{d}(\frac{1}{9d}-\frac{1}{9d+nd})$=$\frac{1}{{d}^{2}}(\frac{1}{9}-\frac{1}{n+9})$.
∴$\frac{1}{{d}^{2}}=1$,解得d=±1.
∴数列{an}的公差d=1或d=-1.
点评 本题考查数列递推式,考查了等差数列的通项公式,考查等比数列的性质,训练了裂项相消法求数列的和,是中档题.