The sum of three numbers of AP is 30 and the sum of their squares is 398 .Find the numbers
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}.
Let's assume the three numbers in the arithmetic progression (AP) are a-d, a, and a+d, where a is the middle term and d is the common difference.
Given:
(a-d) + a + (a+d) = 30 (Sum of three numbers is 30)
(a-d)^2 + a^2 + (a+d)^2 = 398 (Sum of their squares is 398)
We can solve these two equations simultaneously to find the values of a and d.
First, simplify the expressions:
3a = 30 (simplify the first equation by combining like terms)
a = 10 (divide both sides by 3)
Now substitute the value of a in the second equation:
(a-d)^2 + a^2 + (a+d)^2 = 398
(10-d)^2 + 10^2 + (10+d)^2 = 398
(100 - 20d + d^2) + 100 + (100 + 20d + d^2) = 398
200 + 2d^2 = 398
2d^2 = 198
d^2 = 99
d = ±√(99)
d = ±3√11
Since d is the common difference and we assume it is positive, d = 3√11.
Now substitute the value of d in the first equation to find a-d and a+d:
(a-d) + a + (a+d) = 30
10 - 3√11 + 10 + 10 + 3√11 = 30
20 + 3√11 = 30
3√11 = 10
√11 = 10/3
11 = (10/3)^2
11 = 100/9
99 = 100/9
99 ≠ 100/9
Since we have reached a contradiction, the given information is inconsistent. There are no three numbers in the arithmetic progression whose sum is 30 and the sum of their squares is 398.