Question

three numbers are in an arithmetic progression. Their sum is 3 and the sum of their squares is 11. what are the three numbers

Answer 1

The three numbers are 12, 18 and 24

Arithmetic progression

Let the 3 number in arithmetic progression be:

a-d, d, a+d ...

If their sum is 3, then;

a-d+d+a+d = 3

2a + d = 3 ........... 1

If the sum of their squares is 11, then;

(a-d)² + d² + (a+d)² = 11

a²-2ad+d²+d²+a²+2ad+d²  11

2a²+3d² = 11 ....... 2

Solving the equations simultaneously, d = 6 and a = 12

First-term = 12

second term = 18

Thirs term = 24

Hence the three numbers are 12, 18 and 24

Hope this helps you!!!!!! :D

Answer 2

Answer:

-3, 1, -1 or -1, 1, 3

Step-by-step explanation:

We can create the equation of x + x+d + x+2d = 3 or use the sum formula to end up with x = 1-d.

We can then calculate d by using 11 and creating an equation like this.

x^2 + (x+d)^2 + (x+2d)^2 = 11 -> (1-d)^2 + (1)^2 + (1+d)^2 =11 which ends up with d equaling -2 or 2

If we plug d into the sum formula we get

(2x + 2 * 2)/2 *3 =3

and

(2x + 2 * -2)/2 *3 =3

This leaves us with starting terms -1 and 3. We can then finish the arithmetic sequence for each case by just using d.