To determine how many terms of the arithmetic series [tex]\(2 + 4 + 6 + \ldots\)[/tex] sum up to 420, we need to use the formula for the sum of an arithmetic series:
[tex]\[
S_n = \frac{n}{2} (2a + (n - 1)d)
\][/tex]
where [tex]\( S_n \)[/tex] is the sum of the first [tex]\( n \)[/tex] terms, [tex]\( a \)[/tex] is the first term, [tex]\( d \)[/tex] is the common difference, and [tex]\( n \)[/tex] is the number of terms.
Given:
- The sum needed, [tex]\( S_n = 420 \)[/tex]
- The first term, [tex]\( a = 2 \)[/tex]
- The common difference, [tex]\( d = 2 \)[/tex]
We need to find the value of [tex]\( n \)[/tex] where [tex]\( S_n = 420 \)[/tex]. Let’s plug in the values into the sum formula:
[tex]\[
420 = \frac{n}{2} (2 \times 2 + (n - 1) \times 2)
\][/tex]
Simplify the equation:
[tex]\[
420 = \frac{n}{2} (4 + 2n - 2)
\][/tex]
[tex]\[
420 = \frac{n}{2} (2 + 2n)
\][/tex]
[tex]\[
420 = \frac{n}{2} \times 2(n + 1)
\][/tex]
[tex]\[
420 = n(n + 1)
\][/tex]
This simplifies to:
[tex]\[
n^2 + n - 420 = 0
\][/tex]
Now, we solve for [tex]\( n \)[/tex] using the quadratic formula [tex]\( n = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -420 \)[/tex]:
1. Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = 1^2 - 4(1)(-420) = 1 + 1680 = 1681
\][/tex]
2. Solve for [tex]\( n \)[/tex]:
[tex]\[
n = \frac{{-1 \pm \sqrt{1681}}}{2}
\][/tex]
[tex]\[
\sqrt{1681} = 41
\][/tex]
[tex]\[
n = \frac{{-1 \pm 41}}{2}
\][/tex]
This gives us two solutions:
[tex]\[
n_1 = \frac{{-1 + 41}}{2} = \frac{40}{2} = 20
\][/tex]
[tex]\[
n_2 = \frac{{-1 - 41}}{2} = \frac{{-42}}{2} = -21
\][/tex]
Since [tex]\( n \)[/tex] must be a positive integer, we take [tex]\( n = 20 \)[/tex].
Therefore, the number of terms of the series [tex]\(2 + 4 + 6 + \ldots \)[/tex] that amount to 420 is [tex]\( \boxed{20} \)[/tex].
