Answer :
Sure, let's approach and solve each part of the problem step-by-step.
### Given: Arithmetic Series [tex]\(42, 39, 36, 33, \ldots\)[/tex]
#### Part (i): Find the sum of the first 10 terms of the series.
1. Identify the first term ([tex]\(a\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
- The first term [tex]\(a = 42\)[/tex].
- The second term is [tex]\(39\)[/tex]. The common difference [tex]\(d = 39 - 42 = -3\)[/tex].
2. Number of terms ([tex]\(n\)[/tex]):
- We need the sum of the first 10 terms, so [tex]\(n = 10\)[/tex].
3. Sum of the first [tex]\(n\)[/tex] terms formula:
The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series is given by:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
4. Substitute [tex]\(a = 42\)[/tex], [tex]\(d = -3\)[/tex], and [tex]\(n = 10\)[/tex] into the formula:
[tex]\[ S_{10} = \frac{10}{2} \left(2 \cdot 42 + (10-1) \cdot (-3)\right) \][/tex]
[tex]\[ S_{10} = 5 \left(84 - 27\right) \][/tex]
[tex]\[ S_{10} = 5 \times 57 \][/tex]
[tex]\[ S_{10} = 285 \][/tex]
Thus, the sum of the first 10 terms of the series is [tex]\(285\)[/tex].
#### Part (ii): If the sum of the first [tex]\(n\)[/tex] terms of the series is 315, find the possible values of [tex]\(n\)[/tex].
1. Given:
[tex]\[ S_n = 315 \][/tex]
2. Use the sum formula:
[tex]\[ 315 = \frac{n}{2} \left(2 \cdot 42 + (n-1) \cdot (-3)\right) \][/tex]
3. Simplify inside the parentheses:
[tex]\[ 315 = \frac{n}{2} (84 - 3n + 3) \][/tex]
[tex]\[ 315 = \frac{n}{2} (87 - 3n) \][/tex]
[tex]\[ 315 = \frac{87n - 3n^2}{2} \][/tex]
4. Multiply both sides by 2 to clear the fraction:
[tex]\[ 630 = 87n - 3n^2 \][/tex]
[tex]\[ 3n^2 - 87n + 630 = 0 \][/tex]
5. Solve this quadratic equation:
[tex]\[ 3n^2 - 87n + 630 = 0 \][/tex]
Using 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]:
6. Calculate the discriminant ([tex]\(\Delta\)[/tex]) and solve for [tex]\(n\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = (-87)^2 - 4 \cdot 3 \cdot 630 \][/tex]
[tex]\[ \Delta = 7569 - 7560 \][/tex]
[tex]\[ \Delta = 9 \][/tex]
Substituting back into the quadratic formula:
[tex]\[ n = \frac{87 \pm \sqrt{9}}{6} \][/tex]
[tex]\[ n = \frac{87 \pm 3}{6} \][/tex]
7. Calculate the possible values of [tex]\(n\)[/tex]:
[tex]\[ n = \frac{90}{6} = 15 \quad \text{and} \quad n = \frac{84}{6} = 14 \][/tex]
Thus, the possible values of [tex]\(n\)[/tex] are [tex]\(14\)[/tex] and [tex]\(15\)[/tex].
#### Part (iii): Explain the double values of [tex]\(n\)[/tex] obtained in (ii).
The occurrence of two values for [tex]\(n\)[/tex] is a result of solving a quadratic equation. In this particular problem, we derived a quadratic equation from the sum of the arithmetic series:
[tex]\[ 3n^2 - 87n + 630 = 0 \][/tex]
This equation represents a parabola, and parabolas often intersect a horizontal line (representing the sum of 315) at two points. Hence, we get two potential solutions for [tex]\(n\)[/tex]. Each solution represents a valid sequence length ([tex]\(n\)[/tex]) that satisfies the condition [tex]\(S_n = 315\)[/tex]. Therefore, the double values of [tex]\(n\)[/tex] occur due to the nature of quadratic equations, which can have up to two real roots.
### Given: Arithmetic Series [tex]\(42, 39, 36, 33, \ldots\)[/tex]
#### Part (i): Find the sum of the first 10 terms of the series.
1. Identify the first term ([tex]\(a\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
- The first term [tex]\(a = 42\)[/tex].
- The second term is [tex]\(39\)[/tex]. The common difference [tex]\(d = 39 - 42 = -3\)[/tex].
2. Number of terms ([tex]\(n\)[/tex]):
- We need the sum of the first 10 terms, so [tex]\(n = 10\)[/tex].
3. Sum of the first [tex]\(n\)[/tex] terms formula:
The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series is given by:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
4. Substitute [tex]\(a = 42\)[/tex], [tex]\(d = -3\)[/tex], and [tex]\(n = 10\)[/tex] into the formula:
[tex]\[ S_{10} = \frac{10}{2} \left(2 \cdot 42 + (10-1) \cdot (-3)\right) \][/tex]
[tex]\[ S_{10} = 5 \left(84 - 27\right) \][/tex]
[tex]\[ S_{10} = 5 \times 57 \][/tex]
[tex]\[ S_{10} = 285 \][/tex]
Thus, the sum of the first 10 terms of the series is [tex]\(285\)[/tex].
#### Part (ii): If the sum of the first [tex]\(n\)[/tex] terms of the series is 315, find the possible values of [tex]\(n\)[/tex].
1. Given:
[tex]\[ S_n = 315 \][/tex]
2. Use the sum formula:
[tex]\[ 315 = \frac{n}{2} \left(2 \cdot 42 + (n-1) \cdot (-3)\right) \][/tex]
3. Simplify inside the parentheses:
[tex]\[ 315 = \frac{n}{2} (84 - 3n + 3) \][/tex]
[tex]\[ 315 = \frac{n}{2} (87 - 3n) \][/tex]
[tex]\[ 315 = \frac{87n - 3n^2}{2} \][/tex]
4. Multiply both sides by 2 to clear the fraction:
[tex]\[ 630 = 87n - 3n^2 \][/tex]
[tex]\[ 3n^2 - 87n + 630 = 0 \][/tex]
5. Solve this quadratic equation:
[tex]\[ 3n^2 - 87n + 630 = 0 \][/tex]
Using 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]:
6. Calculate the discriminant ([tex]\(\Delta\)[/tex]) and solve for [tex]\(n\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = (-87)^2 - 4 \cdot 3 \cdot 630 \][/tex]
[tex]\[ \Delta = 7569 - 7560 \][/tex]
[tex]\[ \Delta = 9 \][/tex]
Substituting back into the quadratic formula:
[tex]\[ n = \frac{87 \pm \sqrt{9}}{6} \][/tex]
[tex]\[ n = \frac{87 \pm 3}{6} \][/tex]
7. Calculate the possible values of [tex]\(n\)[/tex]:
[tex]\[ n = \frac{90}{6} = 15 \quad \text{and} \quad n = \frac{84}{6} = 14 \][/tex]
Thus, the possible values of [tex]\(n\)[/tex] are [tex]\(14\)[/tex] and [tex]\(15\)[/tex].
#### Part (iii): Explain the double values of [tex]\(n\)[/tex] obtained in (ii).
The occurrence of two values for [tex]\(n\)[/tex] is a result of solving a quadratic equation. In this particular problem, we derived a quadratic equation from the sum of the arithmetic series:
[tex]\[ 3n^2 - 87n + 630 = 0 \][/tex]
This equation represents a parabola, and parabolas often intersect a horizontal line (representing the sum of 315) at two points. Hence, we get two potential solutions for [tex]\(n\)[/tex]. Each solution represents a valid sequence length ([tex]\(n\)[/tex]) that satisfies the condition [tex]\(S_n = 315\)[/tex]. Therefore, the double values of [tex]\(n\)[/tex] occur due to the nature of quadratic equations, which can have up to two real roots.
