If the opposite number of a number is non positive, then the number must be: A, positive B, non positive C, non negative D, negative

If the opposite number of a number is non positive, then the number must be: A, positive B, non positive C, non negative D, negative

Non negative numbers are right

It is known that three real numbers are equal difference sequence. The number obtained by multiplying the smallest number by 2, the largest number plus 7 is equal ratio sequence, and their product is 1000
It is known that three real numbers form an arithmetic sequence. The number obtained by multiplying the smallest number of the three numbers by 2 and the largest number plus 7 will form an arithmetic sequence, and their product is 1000. The tolerance of arithmetic sequence can be obtained

Let three real numbers be A-D, a, a + D, then 2 (A-D), a, a + D + 7 become an equal ratio sequence, ∩ a ^ 2 = 2 (A-D) * (a + D + 7) and 2 (A-D) * a * (a + D + 7) = 1000, that is, a ^ 3 = 1000, so a = 10, so 10 ^ 2 = 2 (10-d) * (10 + D + 7) is reduced to d ^ 2 + 14d-120 = 0, and the solution is d = 8 or - 15, so the tolerance is 8 or - 15

For the negative real number a, the numbers 4A + 3, 7a + 7, a ^ 2 + 8A + 3 form the arithmetic sequence in turn
(1) Find the value of A;
(2) If the sequence {an} satisfies an + 1 = a ^ (n + 1) - 2An (n ∈ n +), A1 = M
① Prove that the sequence {an / (- 2) ^ n} is the general term formula of arithmetic sequence {an}
(3) Under the condition of (2), if for any n ∈ n +, the inequality a (2n + 1)

1.
(4a+3)+(a^2+8a+3)=2(7a+7)
a^2-2a-8=0
(a+2)(a-4)=0
a=-2.
two
①
a(n+1)=a^(n+1)-2an=(-2)^(n+1)-2an
a(n+1)=(-2)^(n+1)-2an
a(n+1)/(-2)^(n+1)=1-2an/(-2)^(n+1)
=1+an/(-2)^n
a(n+1)/(-2)^(n+1)-an/(-2)^n=1
So an / (- 2) ^ n is an arithmetic sequence with tolerance of 1 and first term of A1 / (- 2) = - M / 2;
②
Because a (n + 1) / (- 2) ^ (n + 1) - an / (- 2) ^ n = 1
therefore
an/(-2)^n=a1/(-2)+(n-1)*1
=-m/2+n-1
an=(-m/2+n-1)(-2)^n.
three
a(2n+1)=[-m/2+(2n+1)-1](-2)^(2n+1)
=[-m/2+2n](-2)^(2n+1)
a(2n-1)=[-m/2+(2n-1)-1](-2)^(2n-1)
=[-m/2+2n-2](-2)^(2n-1)
a(2n+1)＜a(2n-1)
[-m/2+2n](-2)^(2n+1)＜[-m/2+2n-2](-2)^(2n-1)
[-2m+8n](-2)^(2n-1)＜[-m/2+2n-2](-2)^(2n-1)
-2m+8n＞-m/2+2n-2
-3m/2+6n+2＞0
n＞(3m/2-2)/6
Because n ≥ 1, as long as (3m / 2-2) / 6 < 1
therefore
3m/2-2＜6
m＜16/3.

If you multiply any two real numbers - 5, - 3,4, the largest product is______ .
If you multiply any two real numbers - 5, - 3,4, the largest product is（

If you multiply any two real numbers - 5, - 3,4, the largest product is (15)