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:
-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.
