The sum of three numbers in 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)
Their sum is given as 30, so (a-d) + a + (a+d) = 30
Simplifying, we get 3a = 30, or a = 10
Now we use the second condition, the sum of their squares:
(a-d)^2 + a^2 + (a+d)^2 = 398
Simplifying using a=10, we get:
3a^2 + 2d^2 = 398
Substituting a=10, we get:
30 + 2d^2 = 398
Simplifying, we get:
d^2 = 184
Taking the square root, we get:
d = ±√184 = ±2√46
Since the numbers are in AP, we know that d must be positive (otherwise we would have a negative number in the sequence). Therefore, d = 2√46
Substituting this value of d into our expression for the numbers in AP, we get:
a-d = 10 - 2√46
a = 10
a+d = 10 + 2√46
Therefore, the three numbers are:
10 - 2√46, 10, 10 + 2√46
Let's assume the three numbers in the arithmetic progression are a-d, a, and a+d.
According to the problem, the sum of these three numbers is 30:
(a-d) + a + (a+d) = 30
Simplifying the equation, we get:
3a = 30
Dividing both sides by 3, we find:
a = 10
Now, let's find the value of d.
Squaring the numbers in the progression, we get:
(a-d)^2 + a^2 + (a+d)^2 = 398
Expanding the equation, we have:
(a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 398
Combine like terms:
3a^2 + 2d^2 = 398
Substituting the value of a, we have:
3(10)^2 + 2d^2 = 398
Simplifying the equation:
300 + 2d^2 = 398
Subtracting 300 from both sides:
2d^2 = 98
Divide both sides by 2:
d^2 = 49
Taking the square root of both sides, we find:
d = ±7
Therefore, the three numbers in the arithmetic progression are:
a-d = 10-7 = 3
a = 10
a+d = 10+7 = 17
So, the three numbers are 3, 10, and 17.