Answer :
Let's solve the problem step-by-step.
### Part (a): Calculate the common difference of the series.
1. Define the terms:
- Let [tex]\( a \)[/tex] be the first term of the arithmetic series.
- Let [tex]\( d \)[/tex] be the common difference of the series.
2. Given information:
- The third term of the arithmetic series, [tex]\( a + 2d \)[/tex], is 70.
- The sum of the first 10 terms of the series, [tex]\( S_{10} \)[/tex], is 450.
3. Use the given information to set up equations:
- The third term equation is:
[tex]\[ a + 2d = 70 \][/tex]
- The formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic series is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
Plugging in [tex]\( n = 10 \)[/tex] and [tex]\( S_{10} = 450 \)[/tex]:
[tex]\[ 450 = \frac{10}{2} \left(2a + 9d\right) \][/tex]
Simplify to get:
[tex]\[ 450 = 5 \left(2a + 9d\right) \][/tex]
[tex]\[ 2a + 9d = 90 \][/tex]
4. Solve the system of linear equations:
- We have two equations:
1. [tex]\( a + 2d = 70 \)[/tex]
2. [tex]\( 2a + 9d = 90 \)[/tex]
- Solve equation 1 for [tex]\( a \)[/tex]:
[tex]\[ a = 70 - 2d \][/tex]
- Substitute [tex]\( a = 70 - 2d \)[/tex] into equation 2:
[tex]\[ 2(70 - 2d) + 9d = 90 \][/tex]
Simplify:
[tex]\[ 140 - 4d + 9d = 90 \][/tex]
[tex]\[ 140 + 5d = 90 \][/tex]
[tex]\[ 5d = 90 - 140 \][/tex]
[tex]\[ 5d = -50 \][/tex]
[tex]\[ d = -10 \][/tex]
- Find [tex]\( a \)[/tex] using [tex]\( a = 70 - 2d \)[/tex]:
[tex]\[ a = 70 - 2(-10) \][/tex]
[tex]\[ a = 70 + 20 \][/tex]
[tex]\[ a = 90 \][/tex]
So, the first term [tex]\( a = 90 \)[/tex] and the common difference [tex]\( d = -10 \)[/tex].
### Part (b): Find the set of possible values of [tex]\( n \)[/tex] when [tex]\( S_n \geq 350 \)[/tex].
1. Given inequality:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \geq 350 \][/tex]
Substitute [tex]\( a = 90 \)[/tex] and [tex]\( d = -10 \)[/tex] into the inequality:
[tex]\[ \frac{n}{2} \left(2(90) + (n-1)(-10)\right) \geq 350 \][/tex]
Simplify inside the parentheses:
[tex]\[ \frac{n}{2} \left(180 - 10(n-1)\right) \geq 350 \][/tex]
[tex]\[ \frac{n}{2} \left(180 - 10n + 10\right) \geq 350 \][/tex]
[tex]\[ \frac{n}{2} \left(190 - 10n\right) \geq 350 \][/tex]
[tex]\[ \frac{n}{2} \cdot 190 - 5n^2 \geq 350 \][/tex]
[tex]\[ 95n - 5n^2 \geq 350 \][/tex]
2. Rearrange the inequality:
[tex]\[ -5n^2 + 95n - 350 \geq 0 \][/tex]
3. Solve the quadratic inequality:
- First, solve the quadratic equation [tex]\( -5n^2 + 95n - 350 = 0 \)[/tex]:
Use the quadratic formula [tex]\( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = -5, \quad b = 95, \quad c = -350 \][/tex]
[tex]\[ n = \frac{-95 \pm \sqrt{95^2 - 4(-5)(-350)}}{2(-5)} \][/tex]
[tex]\[ n = \frac{-95 \pm \sqrt{9025 - 7000}}{-10} \][/tex]
[tex]\[ n = \frac{-95 \pm \sqrt{2025}}{-10} \][/tex]
[tex]\[ n = \frac{-95 \pm 45}{-10} \][/tex]
- Solving for [tex]\( n \)[/tex]:
Case 1:
[tex]\[ n = \frac{-95 + 45}{-10} = \frac{-50}{-10} = 5 \][/tex]
Case 2:
[tex]\[ n = \frac{-95 - 45}{-10} = \frac{-140}{-10} = 14 \][/tex]
4. Analyze the quadratic inequality:
- The quadratic expression [tex]\( -5n^2 + 95n - 350 \)[/tex] opens downwards (since the coefficient of [tex]\( n^2 \)[/tex] is negative).
- The roots are [tex]\( n = 5 \)[/tex] and [tex]\( n = 14 \)[/tex].
Therefore, the inequality [tex]\( -5n^2 + 95n - 350 \geq 0 \)[/tex] is satisfied for values of [tex]\( n \)[/tex] between the roots, inclusive:
[tex]\[ 5 \leq n \leq 14 \][/tex]
Thus, the set of possible values of [tex]\( n \)[/tex] for which [tex]\( S_n \geq 350 \)[/tex] is:
[tex]\[ n \in \{5, 6, 7, 8, 9, 10, 11, 12, 13, 14\} \][/tex]
### Part (a): Calculate the common difference of the series.
1. Define the terms:
- Let [tex]\( a \)[/tex] be the first term of the arithmetic series.
- Let [tex]\( d \)[/tex] be the common difference of the series.
2. Given information:
- The third term of the arithmetic series, [tex]\( a + 2d \)[/tex], is 70.
- The sum of the first 10 terms of the series, [tex]\( S_{10} \)[/tex], is 450.
3. Use the given information to set up equations:
- The third term equation is:
[tex]\[ a + 2d = 70 \][/tex]
- The formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic series is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
Plugging in [tex]\( n = 10 \)[/tex] and [tex]\( S_{10} = 450 \)[/tex]:
[tex]\[ 450 = \frac{10}{2} \left(2a + 9d\right) \][/tex]
Simplify to get:
[tex]\[ 450 = 5 \left(2a + 9d\right) \][/tex]
[tex]\[ 2a + 9d = 90 \][/tex]
4. Solve the system of linear equations:
- We have two equations:
1. [tex]\( a + 2d = 70 \)[/tex]
2. [tex]\( 2a + 9d = 90 \)[/tex]
- Solve equation 1 for [tex]\( a \)[/tex]:
[tex]\[ a = 70 - 2d \][/tex]
- Substitute [tex]\( a = 70 - 2d \)[/tex] into equation 2:
[tex]\[ 2(70 - 2d) + 9d = 90 \][/tex]
Simplify:
[tex]\[ 140 - 4d + 9d = 90 \][/tex]
[tex]\[ 140 + 5d = 90 \][/tex]
[tex]\[ 5d = 90 - 140 \][/tex]
[tex]\[ 5d = -50 \][/tex]
[tex]\[ d = -10 \][/tex]
- Find [tex]\( a \)[/tex] using [tex]\( a = 70 - 2d \)[/tex]:
[tex]\[ a = 70 - 2(-10) \][/tex]
[tex]\[ a = 70 + 20 \][/tex]
[tex]\[ a = 90 \][/tex]
So, the first term [tex]\( a = 90 \)[/tex] and the common difference [tex]\( d = -10 \)[/tex].
### Part (b): Find the set of possible values of [tex]\( n \)[/tex] when [tex]\( S_n \geq 350 \)[/tex].
1. Given inequality:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \geq 350 \][/tex]
Substitute [tex]\( a = 90 \)[/tex] and [tex]\( d = -10 \)[/tex] into the inequality:
[tex]\[ \frac{n}{2} \left(2(90) + (n-1)(-10)\right) \geq 350 \][/tex]
Simplify inside the parentheses:
[tex]\[ \frac{n}{2} \left(180 - 10(n-1)\right) \geq 350 \][/tex]
[tex]\[ \frac{n}{2} \left(180 - 10n + 10\right) \geq 350 \][/tex]
[tex]\[ \frac{n}{2} \left(190 - 10n\right) \geq 350 \][/tex]
[tex]\[ \frac{n}{2} \cdot 190 - 5n^2 \geq 350 \][/tex]
[tex]\[ 95n - 5n^2 \geq 350 \][/tex]
2. Rearrange the inequality:
[tex]\[ -5n^2 + 95n - 350 \geq 0 \][/tex]
3. Solve the quadratic inequality:
- First, solve the quadratic equation [tex]\( -5n^2 + 95n - 350 = 0 \)[/tex]:
Use the quadratic formula [tex]\( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = -5, \quad b = 95, \quad c = -350 \][/tex]
[tex]\[ n = \frac{-95 \pm \sqrt{95^2 - 4(-5)(-350)}}{2(-5)} \][/tex]
[tex]\[ n = \frac{-95 \pm \sqrt{9025 - 7000}}{-10} \][/tex]
[tex]\[ n = \frac{-95 \pm \sqrt{2025}}{-10} \][/tex]
[tex]\[ n = \frac{-95 \pm 45}{-10} \][/tex]
- Solving for [tex]\( n \)[/tex]:
Case 1:
[tex]\[ n = \frac{-95 + 45}{-10} = \frac{-50}{-10} = 5 \][/tex]
Case 2:
[tex]\[ n = \frac{-95 - 45}{-10} = \frac{-140}{-10} = 14 \][/tex]
4. Analyze the quadratic inequality:
- The quadratic expression [tex]\( -5n^2 + 95n - 350 \)[/tex] opens downwards (since the coefficient of [tex]\( n^2 \)[/tex] is negative).
- The roots are [tex]\( n = 5 \)[/tex] and [tex]\( n = 14 \)[/tex].
Therefore, the inequality [tex]\( -5n^2 + 95n - 350 \geq 0 \)[/tex] is satisfied for values of [tex]\( n \)[/tex] between the roots, inclusive:
[tex]\[ 5 \leq n \leq 14 \][/tex]
Thus, the set of possible values of [tex]\( n \)[/tex] for which [tex]\( S_n \geq 350 \)[/tex] is:
[tex]\[ n \in \{5, 6, 7, 8, 9, 10, 11, 12, 13, 14\} \][/tex]
