根据题意,奇数项比偶数项多一项,偶数项的和乘以公比q再加上a1就是奇数项的和,所以有q*120+a1=100,即a1=100-120q所以只要求出q即可求出a1,奇数项之和=a1*[1-(q²)^(n+1)]/(1-q²),偶数项之和=a1q[1-(q²)...