An is a proportional series, a4a7 521, a3 a8 124, and the common ratio q is an integer, then a10

It should be 512. If 521 is not calculated, the integer q makes the first term a1 and the common ratio is q, then a4=a1*q 3 a7=a1*q 6 a3=a1*q 2 a8=a1*q 7 substituting the above formula a1 2*q 9=-512 a1*(q 2+q 7)=124 appropriate change, then the above equation becomes a3 2*q 5=-512 a3*(1+q 5)=124 a3*(1+2*q 5+q 10)=15376, then q 5=-521 (a3 2) substitution has a3 2-124* a3-512=0 Solve the binary equation to obtain a3=62 66a3=-4, q=-2 or a3=128, q=-1 2. So a3=-4 and q=-2 fit the topic: a10=a1*q 9=a3*q 7=512

I've done this question, the landlord wrote it wrong, it should be a4a7=-512 Note that 521 is a prime number, there is no such integer q, (o) a4*a7=a3*q*a8 q=a3*a8=-512 (1).

a3+a8=124 (2)

So a3 = -512 a8 is substituted into (2) to get -512 a8 + a8 = 124a8) 2-124 (a8) - 512 = 0

a8-128)(a8+4)=0

a8 = 128 or a8 = -4

then a3 = -4 or a3 = 128

a8/a3=q^5=-32,q=-2

or a8 a3=-4 128=-1 32 (because q is an integer, so it is rounded) so q=-2, a8=128

a10=a8*q^2=128*4=512

an} is a proportional series of stool ants, so there is:

a4^2=a3*a5=18

a4^2=18

a6^2=a4*a8=72

a6^2=72

q 2 = root coarse number (a6 2 a4 2).

Root number macro buried (4) = 2

So. q=+/2

a[2]a[9]=a[3]a[8]=-512 a[3]+a[8]=124 a[3]=-4 a[8]=128 or a[3]=128 a[8]=-4

And because the common ratio is an integer, it can only be a[3]=-4 a[8]=128, and the common ratio q=-2

a[12]=128*(-2)^4=2048

a1a9=256a5 2=a1a9=256a5=16 or a5=-16a4+a6=40a5 Round regret dismantling q+a5*q=40a5*(q+1 good q)=401)a5=16q+1 q=5 22q 2+2-5q=0(q-2)(2q-1)=0q1=2 q2=1 22)a5=-16q+1 orange jujube q=-5 22q 2+2+5q=0(q+2)(2q+1)=0q1=-2 q2=- 1 2

a1+a5+a9=8

a1+a1q^4+a1q^8=8………

Song Wangban a4 + a8 + a12 = -27

a1q^3+a1q^7+a1q^11=q^3(a1+a1q^4+a1q^8)=-27…Confessional .........

Get: q 3=-27 Yehu 8

q^3=(-3/2)^3

q=-3/2

a4*a7=-512

So a2*19=-512

a9=a2*q^7

Therefore, a2*a2*q 7=-512, and guess the equation system of a2+a2*q 7=254 to get (1) a2=256, q=-1 2, but the common ratio of the spike or the integer is an integer with a cluster, so this solution is rounded.

2) a2=-2, q=-2, so a12=-2048

a4*a7=a2*q 2*a9 q 2=a2*a9 before solving the equation system: nanofiber a2*a9=-512, a2+a9=254 obtains: a2=256, a9=-2 or a2=-2, a9=256 is an integer, rounding off a2=256, a9=-2

Clear: a9, a2=q, 7=-128

q=-2a12=a9*q^3=2048