Answer :
Let's tackle the question step by step.
### Determining the Number of Terms in the Given Series:
#### Series (a): [tex]\(5 + 8 + 11 + \ldots + 320\)[/tex]
1. Identify the first term ([tex]\(a_1\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
[tex]\[ a_1 = 5, \quad d = 3 \][/tex]
2. Identify the last term ([tex]\(a_n\)[/tex]):
[tex]\[ a_n = 320 \][/tex]
3. Use the formula for the [tex]\(n\)[/tex]-th term of an arithmetic series:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
Rearrange to solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{a_n - a_1}{d} + 1 \][/tex]
4. Substitute the given values:
[tex]\[ n = \frac{320 - 5}{3} + 1 = \frac{315}{3} + 1 = 105 + 1 = 106 \][/tex]
So, the series has 106 terms.
#### Series (c): [tex]\(10 + 13 + 16 + \ldots + 49\)[/tex]
1. Identify the first term ([tex]\(a_1\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
[tex]\[ a_1 = 10, \quad d = 3 \][/tex]
2. Identify the last term ([tex]\(a_n\)[/tex]):
[tex]\[ a_n = 49 \][/tex]
3. Use the formula for the [tex]\(n\)[/tex]-th term of an arithmetic series:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
Rearrange to solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{a_n - a_1}{d} + 1 \][/tex]
4. Substitute the given values:
[tex]\[ n = \frac{49 - 10}{3} + 1 = \frac{39}{3} + 1 = 13 + 1 = 14 \][/tex]
So, the series has 14 terms.
#### Series (b) and (d) cannot be solved as they do not provide sufficient information (e.g., the last term of the series is not given).
### Determine the Common Difference and First Term for the Specific Conditions:
#### (a) [tex]\(10^{\text{th}}\)[/tex] Term [tex]\(= 23\)[/tex] and [tex]\(32^{\text{nd}}\)[/tex] Term [tex]\(= 67\)[/tex]:
1. Use the general formula for the [tex]\(n\)[/tex]-th term of an arithmetic series:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
2. Write two equations using the given terms:
[tex]\[ \begin{cases} a_1 + 9d = 23 \quad & \text{(for the 10th term)} \\ a_1 + 31d = 67 \quad & \text{(for the 32nd term)} \end{cases} \][/tex]
3. Subtract the first equation from the second:
[tex]\[ (a_1 + 31d) - (a_1 + 9d) = 67 - 23 \\ 22d = 44 \\ \][/tex]
4. Solve for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{44}{22} = 2 \][/tex]
5. Substitute [tex]\(d\)[/tex] back into one of the original equations to find [tex]\(a_1\)[/tex]:
[tex]\[ a_1 + 9 \cdot d = 23 \\ a_1 + 9 \cdot 2 = 23 \\ a_1 + 18 = 23 \\ a_1 = 23 - 18 \\ a_1 = 5 \][/tex]
Thus, the first term [tex]\(a_1 = 5\)[/tex] and the common difference [tex]\(d = 2\)[/tex].
#### (b) [tex]\(11^{\text{th}}\)[/tex] Term [tex]\(= 5\)[/tex]
The provided information is incomplete and does not allow us to find the first term and the common difference. More information is required to solve this part.
### Summary:
- Series (a) has 106 terms.
- Series (c) has 14 terms.
- The first term and common difference for the series whose [tex]\(10^{\text{th}}\)[/tex] term is 23 and [tex]\(32^{\text{nd}}\)[/tex] term is 67 are 5 and 2, respectively.
- Insufficient information is provided to solve for series (b) and (d).
### Determining the Number of Terms in the Given Series:
#### Series (a): [tex]\(5 + 8 + 11 + \ldots + 320\)[/tex]
1. Identify the first term ([tex]\(a_1\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
[tex]\[ a_1 = 5, \quad d = 3 \][/tex]
2. Identify the last term ([tex]\(a_n\)[/tex]):
[tex]\[ a_n = 320 \][/tex]
3. Use the formula for the [tex]\(n\)[/tex]-th term of an arithmetic series:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
Rearrange to solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{a_n - a_1}{d} + 1 \][/tex]
4. Substitute the given values:
[tex]\[ n = \frac{320 - 5}{3} + 1 = \frac{315}{3} + 1 = 105 + 1 = 106 \][/tex]
So, the series has 106 terms.
#### Series (c): [tex]\(10 + 13 + 16 + \ldots + 49\)[/tex]
1. Identify the first term ([tex]\(a_1\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
[tex]\[ a_1 = 10, \quad d = 3 \][/tex]
2. Identify the last term ([tex]\(a_n\)[/tex]):
[tex]\[ a_n = 49 \][/tex]
3. Use the formula for the [tex]\(n\)[/tex]-th term of an arithmetic series:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
Rearrange to solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{a_n - a_1}{d} + 1 \][/tex]
4. Substitute the given values:
[tex]\[ n = \frac{49 - 10}{3} + 1 = \frac{39}{3} + 1 = 13 + 1 = 14 \][/tex]
So, the series has 14 terms.
#### Series (b) and (d) cannot be solved as they do not provide sufficient information (e.g., the last term of the series is not given).
### Determine the Common Difference and First Term for the Specific Conditions:
#### (a) [tex]\(10^{\text{th}}\)[/tex] Term [tex]\(= 23\)[/tex] and [tex]\(32^{\text{nd}}\)[/tex] Term [tex]\(= 67\)[/tex]:
1. Use the general formula for the [tex]\(n\)[/tex]-th term of an arithmetic series:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
2. Write two equations using the given terms:
[tex]\[ \begin{cases} a_1 + 9d = 23 \quad & \text{(for the 10th term)} \\ a_1 + 31d = 67 \quad & \text{(for the 32nd term)} \end{cases} \][/tex]
3. Subtract the first equation from the second:
[tex]\[ (a_1 + 31d) - (a_1 + 9d) = 67 - 23 \\ 22d = 44 \\ \][/tex]
4. Solve for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{44}{22} = 2 \][/tex]
5. Substitute [tex]\(d\)[/tex] back into one of the original equations to find [tex]\(a_1\)[/tex]:
[tex]\[ a_1 + 9 \cdot d = 23 \\ a_1 + 9 \cdot 2 = 23 \\ a_1 + 18 = 23 \\ a_1 = 23 - 18 \\ a_1 = 5 \][/tex]
Thus, the first term [tex]\(a_1 = 5\)[/tex] and the common difference [tex]\(d = 2\)[/tex].
#### (b) [tex]\(11^{\text{th}}\)[/tex] Term [tex]\(= 5\)[/tex]
The provided information is incomplete and does not allow us to find the first term and the common difference. More information is required to solve this part.
### Summary:
- Series (a) has 106 terms.
- Series (c) has 14 terms.
- The first term and common difference for the series whose [tex]\(10^{\text{th}}\)[/tex] term is 23 and [tex]\(32^{\text{nd}}\)[/tex] term is 67 are 5 and 2, respectively.
- Insufficient information is provided to solve for series (b) and (d).
