Certainly! Let's determine how many terms are needed in the arithmetic series [tex]\(42, 39, 36, \ldots\)[/tex], to obtain a sum of 315.
1. Identify the series properties:
- The first term ([tex]\(a\)[/tex]) is 42.
- The common difference ([tex]\(d\)[/tex]) is -3.
- The sum we need ([tex]\(S_n\)[/tex]) is 315.
2. Sum formula for arithmetic series:
The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series can be found using the formula:
[tex]\[
S_n = \frac{n}{2} \left(2a + (n-1)d\right)
\][/tex]
3. Set up the equation:
We need to find [tex]\(n\)[/tex] such that the sum [tex]\(S_n\)[/tex] is 315:
[tex]\[
315 = \frac{n}{2} \left(2 \cdot 42 + (n-1)(-3)\right)
\][/tex]
4. Simplify the expression inside the parentheses:
[tex]\[
315 = \frac{n}{2} \left(84 - 3(n-1)\right)
\][/tex]
Simplifying further:
[tex]\[
315 = \frac{n}{2} \left(84 - 3n + 3\right)
\][/tex]
[tex]\[
315 = \frac{n}{2} \left(87 - 3n\right)
\][/tex]
5. Eliminate the fraction by multiplying both sides by 2:
[tex]\[
630 = n (87 - 3n)
\][/tex]
6. Distribute [tex]\(n\)[/tex] and rearrange the equation:
[tex]\[
630 = 87n - 3n^2
\][/tex]
[tex]\[
3n^2 - 87n + 630 = 0
\][/tex]
7. Solve the quadratic equation:
To solve the quadratic equation [tex]\(3n^2 - 87n + 630 = 0\)[/tex], we can use the quadratic formula [tex]\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 3\)[/tex], [tex]\(b = -87\)[/tex], and [tex]\(c = 630\)[/tex].
8. Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = (-87)^2 - 4 \cdot 3 \cdot 630 = 7569 - 7560 = 9
\][/tex]
9. Find the roots:
[tex]\[
n = \frac{-(-87) \pm \sqrt{9}}{2 \cdot 3} = \frac{87 \pm 3}{6}
\][/tex]
This gives us two solutions:
[tex]\[
n = \frac{87 + 3}{6} = \frac{90}{6} = 15
\][/tex]
[tex]\[
n = \frac{87 - 3}{6} = \frac{84}{6} = 14
\][/tex]
So, the two possible values of [tex]\(n\)[/tex] are 14 and 15.
10. Conclusion:
There are two different answers depending on how we interpret the question, either considering [tex]\(n = 14\)[/tex] or [tex]\(n = 15\)[/tex]. Both are valid numbers of terms in the series that provide the sum of 315.
