Certainly! Let's find out how many terms of the series [tex]\(2+4+6+8+\ldots\)[/tex] must be taken for their sum to be 420.
First, note that this is an arithmetic series.
In an arithmetic series, the sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms can be given by the formula:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
where:
- [tex]\(a\)[/tex] is the first term,
- [tex]\(d\)[/tex] is the common difference,
- [tex]\(n\)[/tex] is the number of terms,
- [tex]\(S_n\)[/tex] is the sum of the first [tex]\(n\)[/tex] terms.
For our series:
- The first term [tex]\(a = 2\)[/tex],
- The common difference [tex]\(d = 2\)[/tex],
- The sum we seek [tex]\(S_n = 420\)[/tex].
Let's plug these values into the sum formula to find [tex]\(n\)[/tex]:
[tex]\[ 420 = \frac{n}{2} [2 \cdot 2 + (n-1) \cdot 2] \][/tex]
Simplify inside the brackets:
[tex]\[ 420 = \frac{n}{2} [4 + 2n - 2] \][/tex]
[tex]\[ 420 = \frac{n}{2} [2 + 2n] \][/tex]
[tex]\[ 420 = n(1 + n) \][/tex]
This is a quadratic equation:
[tex]\[ n^2 + n - 420 = 0 \][/tex]
We solve this quadratic equation 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].
[tex]\[ n = \frac{-1 \pm \sqrt{1 + 1680}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm \sqrt{1681}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm 41}{2} \][/tex]
This results in two potential solutions:
[tex]\[ n = \frac{-1 + 41}{2} = \frac{40}{2} = 20 \][/tex]
[tex]\[ n = \frac{-1 - 41}{2} = \frac{-42}{2} = -21 \][/tex]
However, because we are seeking a number of terms, [tex]\(n\)[/tex] must be a positive integer. Therefore, the negative solution doesn't make sense in this context.
Thus, the number of terms needed to make the sum 420 is:
[tex]\[ n = 20 \][/tex]
In summary, you must take 20 terms of the series [tex]\(2, 4, 6, 8, \ldots\)[/tex] in order for the sum to be 420. The negative solution ([tex]\(-21\)[/tex]) is mathematically valid but irrelevant in this context as it doesn't correspond to a physical number of terms.
