The sum of the first X consecutive even numbers is 930. Find X.
10
35
30
41
Explain your solution.
so we want
2 + 4 + 6 + ... = 930
this is an arithmetic series where
a=2 , d=2 , n = ?
n/2[2a + (n-1)d) = 930
n/2[4 + (n-1)(2)] = 930
n^2 + n - 930 = 0
(n+31)(n-30) = 0
n = 30 or n=-31, but n has to be a positive whole number.
there were 30 terms
To find the value of X, which represents the number of consecutive even numbers, we need to set up and solve an equation.
Let's start by finding the formula for the sum of consecutive even numbers. The formula for the sum of an arithmetic series is Sn = (n/2)(a + L), where Sn is the sum, n is the number of terms, a is the first term, and L is the last term.
In this case, the first term is 2, and the last term is the Xth even number, which we can find by multiplying X by 2. So, L = 2X.
Now, we can plug the values into the formula and set it equal to 930:
930 = (X/2)(2 + 2X)
Simplifying the equation:
930 = X(1 + X)
Expanding the equation:
930 = X + X^2
Rearranging the equation to a quadratic form:
X^2 + X - 930 = 0
Now, we can solve the quadratic equation. This can be done by factoring, using the quadratic formula, or by completing the square. In this case, completing the square would be the most effective method. However, I will use the quadratic formula for simplicity:
X = (-b ± √(b^2 - 4ac)) / 2a
Applying the quadratic formula with a = 1, b = 1, and c = -930:
X = (-1 ± √(1^2 - 4(1)(-930))) / (2(1))
Simplifying the equation:
X = (-1 ± √(1 + 3720)) / 2
X = (-1 ± √3721) / 2
X = (-1 ± 61) / 2
Now we have two possible values for X:
X1 = (-1 + 61) / 2 = 60/2 = 30
X2 = (-1 - 61) / 2 = -62/2 = -31
Since the number of consecutive even numbers cannot be negative, we disregard X2 and conclude that X = 30.
Therefore, the correct answer is 30.