Question

10．设公比q（q＞0）的等比数列{a_n}的前n项和S_n，若S₂=3a₂+2，S₄=3a₄+2，求公比q．

Answer 1

分析 由题意可得q和a₁的方程组，解方程组可得．

解答 解：由题意可得$\left\{\begin{array}{l}{{S}_{2}={a}_{1}+{a}_{2}=3{a}_{2}+2}\\{{S}_{4}={a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=3{a}_{4}+2}\end{array}\right.$，
∴$\left\{\begin{array}{l}{{a}_{1}-2{a}_{2}=2}\\{{a}_{1}+{a}_{2}+{a}_{3}-2{a}_{4}=2}\end{array}\right.$，即$\left\{\begin{array}{l}{{a}_{1}-2{a}_{1}q=2}\\{{a}_{1}+{a}_{1}q+{a}_{1}{q}^{2}-2{a}_{1}{q}^{3}=2}\end{array}\right.$，
解得q=$\frac{3}{2}$或q=-1，由q＞0可得q=$\frac{3}{2}$

点评 本题考查等比数列的通项公式，属基础题．