等比数列a4-a2=a2+a3=12求a1和q 求s6

a4-a2=a2(q²-1)=12≠0
q²≠1 q≠1且q≠-1
a4-a2=a2+a3
a2q²-a2=a2+a2q
等式两边同除以a2
q²-1=1+q
q²-q-2=0
(q-2)(q+1)=0
q=2或q=-1(舍去)
a4-a2=a2(q²-1)=3a2=12 a2=4 a1=a2/q=4/2=2
S6=a1(q^6 -1)/(q-1)
=2×(2^6 -1)/(2-1)
=2^7 -2
=128-2
=126a1呢？你没注意看啊，倒数第6行写的很清楚： a1=a2/q=4/2=2