Three numbers are in arithmetic progression. Their sum is 3 and the sum of their squares is 11. What are the three numbers? I got one of the answers which is 5,7,9. What are the other three numbers?

The three numbers are -1, 1, and 3

Explanation:

The three terms sum to

first + second + third = (x)+(x+d)+(x+2d) = 3x+3d

We're told this sum is 3, so,

3x+3d = 3

3(x+d) = 3

x+d = 3/3

x+d = 1

x = 1-d

The three terms x, x+d, x+2d have the squares of x^2, (x+d)^2, and (x+2d)^2 in that order.

Add up those squares and set the sum equal to 11

x^2 + (x+d)^2 + (x+2d)^2 = 11

Then plug in x = 1-d and solve for d.

x^2 + (x+d)^2 + (x+2d)^2 = 11

(1-d)^2 + (1-d+d)^2 + (1-d+2d)^2 = 11

(1-d)^2 + (1)^2 + (1+d)^2 = 11

(1-2d+d^2) + 1 + (1+2d+d^2) = 11

2d^2+3 = 11

2d^2 = 11-3

2d^2 = 8

d^2 = 8/2

d^2 = 4

d = sqrt(4) or d = -sqrt(4)

d = 2 or d = -2

If d = 2, then,

One possible answer is to have the sequence -1, 1, 3

Summing the original terms gets us -1+1+3 = 3

Summing their squares gets us (-1)^2 + (1)^2 + (3)^2 = 1+1+9 = 11

We've confirmed this answer.

If d = -2, then,

We get the same terms, just in the reverse order.

The sequence overall is still the same when we write the numbers from smallest to largest.

So there's only one solution and it's {-1, 1, 3}

< 6 cm to > 26 cm i believe

step-by-step explanation:

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: