Knowing that three numbers are in an equal proportional series, their sum is 26, and the product is

Since the product is negative, their ratio must be negative, and only if the middle number is negative, the product can be negative; Let these three numbers be x, y, z, and the ratio is t; And because sum is a positive number, we can know that x>0, y<0, z>0 (Note: |x|+|z|>|y|）;column equation: x+xt+xt 2=26; x^3*t^3=-512;Solution:

xt=-8；From this we can find x=2, t=-4, so y=-8, z=32. These three numbers are 2, -8, 32.

b*b=a*c

a+b+c=26

a*b*c=-512

a*b*c=a*c*b=b*b*b=-512b=-8a+c=34

a*c=64

Let the common ratio be d, then b d+b*d=34

d = -4 or -1 4

These three numbers are 2,-8,32 or 32,-8,2

Let these three numbers be 1 k*x, x, kx

Their product is -512: x 3=-512 x = -88 + 1 k*x + kx = 26 4k 2 + 17k + 4 = 0 (4k + 1)(k + 4) = 0

k=-1 4 or -4

So: The three numbers are: 2, -8, 32

Let the three numbers be a p, a, ap

a／p+a+ap =26

a／p×a×ap=-512

So a=-8

p = -4 or p = -1 4

So these three numbers are 2, -8, 32

Assuming these three numbers are x q, x, xq, then x q+x+xq=26, and x 3=-512, we get x= -8, and substituting the previous equation gives q= -4

So the three numbers are 2, -8, 32

Actually, I didn't know it at first, and then I was cautious and copied and pasted, haha, this time I'll grind my limbs.

This question examines the knowledge of functions.

Let the number of the first spine hall be a1, the common ratio is q, according to the meaning of the title: a1 + a1q + a1q 2 = 26, a1xa1qxa1q 2 = 216, solution: a1 = 2, a1q = 6, a1q 2 = 18

Or, a1 = 18, a1q = 6, a1q 2 = 2So the three numbers are: 2, 6, 18

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As long as these three numbers are ax, ax2, ax3 (x is the 1st power of x, x2 is the 2nd power of x, x3 is the 3rd power of the medium module x), a+ax+ax2=21, a3x3=64, and the unselling file has to sell and match 1:x=, a=16 2: x=4, a=1

Let the middle number brother Lu Min be x, then the three numbers are x q, x, xq, multiplied by the three numbers, x3=216, x=6, then q=3 The three envy branches are 2 6 18

Let the three numbers be x, y, z

x+y+z=14 1

x*y*z=-216 2

z/y=y/x 3

There are 3 got. z=(y*y)/x

Bring in 2y=-6

Z=18, x=2 with 1 and 2

The three numbers are 2, -6, and 18

Let these three numbers be x q, x, xq respectively

So x q*x*xq=-216

x^3=-216

x=-6, so these three numbers are -6 q, -6, -6q, because -6 q+(-6)+(6*q)=14

6)*(1/q+1+q)=14

So 1 q+1+q=-7 3

So 1 q+q+10 3=0

So 3q 2 + 10q + 3 = 0

3q+1)(q+3)=0

q = -1 3 or q = -3

So these three numbers are.

6 (-1 3), -6, -6*(-1 3) or -6 (-3), -6, -6*(-3).

That is, these three numbers are or

If the product of the three numbers is proportional to the serial state, and their product 216 is set to be x, and the common ratio is q, then the imaginary number of these three numbers is x q, x, xq (x modulus q) *x*xq = 216

x^3=216

x=6 plus 4 in the middle of the number, then the difference series 2*(x+4)=x q+xq6 q+6q=203+3q 2-10q=0(3q-1)(q-3)=0q=1 3,q=3These three numbers are 2,6,18

Let these three numbers be a, aq, aq (a, q are not 0) a+aq+aq = 13

a·aq·aq²=27

The two formulas are solved by combining the two formulas with absolute preparation: a=1 and q=3 are the middle signs