Final answer:
To find the three numbers in the arithmetic progression (AP) with a sum of 33 and a sum of their squares of 461, we can set up equations using the middle number and the common difference. By solving these equations, we find that the three numbers in the AP are 8, 11, and 14.
Step-by-step explanation:
To solve this problem, we can use the fact that the sum of three numbers in an arithmetic progression (AP) is equal to three times the middle number. Let's assume the three numbers are a-d, a, and a+d, where a is the middle number and d is the common difference. We know that (a-d) + a + (a+d) = 33. Simplifying this equation, we get 3a = 33, which means a = 11.
Using this value of a, we can find the common difference (d). We are given that the sum of the squares of the three numbers is 461. So, (a-d)^2 + a^2 + (a+d)^2 = 461. Substituting the value of a and simplifying, we get 3a^2 + 2d^2 = 461.
Since a=11, we can substitute this value into the equation and solve for d. After finding the value of d, we can calculate the three numbers in the AP by subtracting and adding d to a.
Therefore, the three numbers in the AP are 8, 11, and 14.