Answer
- **Problem 4:**
- First term: \(-8\)
- Common difference: \(4\)
- **Problem 5:**
- Common ratio: \(\frac{2}{3}\)
- Number of terms: \(3\)
- **Problem 6:**
- Sum of the first 12 terms: \(360\)
Solution
Let's solve each of the problems step by step.
---
### **Problem 4**
**Given:**
- The sum of the first 5 terms of an arithmetic sequence is 0.
- The fifth term is 8.
**Find:**
- The first term (\( a \))
- The common difference (\( d \))
**Solution:**
1. **Express the fifth term:**
The \( n^{th} \) term of an arithmetic sequence is given by:
\[
T_n = a + (n - 1)d
\]
For the fifth term (\( T_5 \)):
\[
T_5 = a + 4d = 8 \quad \text{(Equation 1)}
\]
2. **Express the sum of the first 5 terms:**
The sum of the first \( n \) terms is:
\[
S_n = \frac{n}{2}(2a + (n - 1)d)
\]
For \( S_5 = 0 \):
\[
S_5 = \frac{5}{2}(2a + 4d) = 0 \implies 2a + 4d = 0 \implies a + 2d = 0 \quad \text{(Equation 2)}
\]
3. **Solve the equations:**
From Equation 2:
\[
a = -2d
\]
Substitute \( a = -2d \) into Equation 1:
\[
-2d + 4d = 8 \implies 2d = 8 \implies d = 4
\]
Then,
\[
a = -2(4) = -8
\]
**Answer:**
- **First term (\( a \))**: \(-8\)
- **Common difference (\( d \))**: \(4\)
---
### **Problem 5**
**Given:**
- The first term of a geometric series is 36.
- The last term is 16.
- The sum of the series is 76.
**Find:**
- The common ratio (\( r \))
- The number of terms (\( n \))
**Solution:**
1. **Express the last term:**
The \( n^{th} \) term of a geometric series is:
\[
T_n = a \cdot r^{n-1}
\]
Given \( T_n = 16 \) and \( a = 36 \):
\[
36 \cdot r^{n-1} = 16 \implies r^{n-1} = \frac{16}{36} = \frac{4}{9} = \left(\frac{2}{3}\right)^2
\]
This implies:
\[
r^{n-1} = \left(\frac{2}{3}\right)^2
\]
Possible solutions:
- \( r = \frac{2}{3} \) and \( n-1 = 2 \) \( \Rightarrow n = 3 \)
- \( r = -\frac{2}{3} \) and \( n-1 = 2 \) \( \Rightarrow n = 3 \) (But we'll check feasibility)
2. **Check the sum for \( r = \frac{2}{3} \) and \( n = 3 \):**
The sum of a geometric series is:
\[
S_n = a \cdot \frac{1 - r^n}{1 - r}
\]
Substituting:
\[
S_3 = 36 \cdot \frac{1 - \left(\frac{2}{3}\right)^3}{1 - \frac{2}{3}} = 36 \cdot \frac{1 - \frac{8}{27}}{\frac{1}{3}} = 36 \cdot \frac{\frac{19}{27}}{\frac{1}{3}} = 36 \cdot \frac{19}{9} = 76
\]
Which matches the given sum.
3. **Check for \( r = -\frac{2}{3} \) and \( n = 3 \):**
Calculating the sum:
\[
S_3 = 36 + 36\left(-\frac{2}{3}\right) + 36\left(\frac{4}{9}\right) = 36 - 24 + 16 = 28 \neq 76
\]
This does not match the given sum.
**Answer:**
- **Common ratio (\( r \))**: \(\frac{2}{3}\)
- **Number of terms (\( n \))**: \(3\)
---
### **Problem 6**
**Given:**
- The sum of the first 4 terms of an arithmetic sequence is 24.
- The 7th term exceeds the 3rd term by 24.
**Find:**
- The sum of the first 12 terms.
**Solution:**
1. **Define variables:**
Let the first term be \( a \) and the common difference be \( d \).
2. **Express the sum of the first 4 terms:**
\[
S_4 = \frac{4}{2}(2a + 3d) = 24 \implies 2(2a + 3d) = 24 \implies 4a + 6d = 24 \implies 2a + 3d = 12 \quad \text{(Equation 1)}
\]
3. **Express the condition on the terms:**
The 7th term (\( T_7 \)) and the 3rd term (\( T_3 \)) are:
\[
T_7 = a + 6d
\]
\[
T_3 = a + 2d
\]
Given:
\[
T_7 - T_3 = 24 \implies (a + 6d) - (a + 2d) = 24 \implies 4d = 24 \implies d = 6
\]
4. **Find the first term \( a \):**
Substitute \( d = 6 \) into Equation 1:
\[
2a + 3(6) = 12 \implies 2a + 18 = 12 \implies 2a = -6 \implies a = -3
\]
5. **Find the sum of the first 12 terms:**
\[
S_{12} = \frac{12}{2}(2a + 11d) = 6(2(-3) + 11 \cdot 6) = 6(-6 + 66) = 6 \times 60 = 360
\]
**Answer:**
- **Sum of the first 12 terms**: \(360\)
---
**Summary of Answers:**
1. **Problem 4:**
- First term (\( a \)): \(-8\)
- Common difference (\( d \)): \(4\)
2. **Problem 5:**
- Common ratio (\( r \)): \(\frac{2}{3}\)
- Number of terms (\( n \)): \(3\)
3. **Problem 6:**
- Sum of the first 12 terms: \(360\)
Reviewed and approved by the UpStudy tutoring team
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