已知｛an｝为等比数列,a4＋a7＝2,a5＊a6＝－8,则a1＋a10＝?

已知｛an｝为等比数列,a4＋a7＝2,a5＊a6＝－8,则a1＋a10＝?

由于是等比数列,a4*a7=a5*a6=-8a4+a7=2解方程有(a4-a7)^2=(a4+a7)^2-4a4*a7=4-4*(-8)=36所以a4=4,a7=-2或a4=-2 a7=4a1:a4=a4:a7所以a1=a4^2/a7,同理,a10=a7^2/a4a4=4,a7=-2时,a1＋a10＝-7a7=4,a4=-2时,a1＋a10＝-7======以下答案可供参考======供参考答案1:31.75a5＊a6=a4＊a7算出q供参考答案2:恭喜你获得自定义头像权限真实头像更能让网友知道你供参考答案3:a4＋a7＝2 a1*q^3+a1q^6=2a5＊a6=-8 a1²q^9=-8 a1q^3 * a1q^6=-8令x1= a1q^3 x2= a1q^6则X1*X2=-8 ,X1+X2=2则X1，X2是方程x²-2x-8=0的两个实数根解得：x1=4，X2=-2先计算 a1q^3=4 则 a1q^6=-2则q^3=-1/2 q^6=1/4a1=-8a1＋a10＝a1+a1*q^3q^6=-8+(-8)(-1/8)=-7如a1q^3=-2 则 a1q^6=4则q^3=-2 q^6=4a1=1a1＋a10＝a1+a1*q^3q^6=1+(-2*4)=-7供参考答案4:因为等比数列a5*a6 = a4*a7(a4+a7)²=a4² + 2*(-8) + a7² = 4 所以 a4² + a7² = 20( a4² + a7²)/a4a7 = (1/q^3 + q^3) = 20/(-8)= -5/2(a4+a7)* (1/q^3 + q^3) = (a1 + a7 + a4 + a10) = a1+a10 + 2所以a1+a10 = 2*(-5/2) -2 = -5-2 = -7

你的回答很对