等比数列a3-a2=4,a2-a1=2,求S5,a3+a3=124,a5*a6=-512,q∈z,求a6,

Question

等比数列a3-a2=4,a2-a1=2,求S5,a3+a3=124,a5*a6=-512,q∈z,求a6,

Answer 1

(a3-a2)= a2*p - a1*p = (a2-a1)*p = 4
又 (a2-a1) = 2,故有:2 *p = 4,所以公比 p = 2
带入a2-a1=2得到：a1*p - a1 = 2 ==> a1*2 - a1=2,==> a1 = 2
故 S5 = a1 *(p^5-1)/(p-1) = 2 * (2^5 -1) /(2-1) = 2*31 = 62
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a3+a3=124 => a3= 62 => a1* p^2 = 62
a5*a6 = -512 => a1*p^4*a1*p^5 = -512 ==>62*62 * p^5=-512.
你这个的题目是不是有问题?计算出来的数很复杂