The sum of three numbers of AP is 30 and the sum of their squares is 398 .Find the numbers

Answer 1

Let the three numbers be a-d, a, and a+d (since they are in AP).
Therefore, their sum is (a-d) + a + (a+d) = 3a, which is given to be 30.
So, a = 10.
Now, the sum of their squares is (a-d)^2 + a^2 + (a+d)^2 = 3a^2 + 2d^2.
Substituting a=10, we get 3(10)^2 + 2d^2 = 398.
Simplifying, we get 2d^2 = 98, or d^2 = 49.
Thus, d can be either 7 or -7.
If d=7, then the numbers are 3, 10, and 17.
If d=-7, then the numbers are 17, 10, and 3.
Therefore, the possible sets of numbers are {3, 10, 17} and {17, 10, 3}.