Answer :
Certainly! Let's solve this step-by-step.
### Part (a)
We need to find the common difference [tex]\(d\)[/tex] and the first term [tex]\(a\)[/tex] of the arithmetic series.
1. Expressions for [tex]\( t_n \)[/tex] and [tex]\( S_n \)[/tex]
- The [tex]\(n\)[/tex]-th term of an arithmetic series is given by:
[tex]\[ t_n = a + (n-1)d \][/tex]
- The sum of the first [tex]\(n\)[/tex] terms [tex]\(S_n\)[/tex] of the series is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
2. Given: [tex]\( S_2 = \frac{2}{3} t_5 \)[/tex]
First, let's express [tex]\(S_2\)[/tex] and [tex]\(t_5\)[/tex]:
- For [tex]\(S_2\)[/tex]:
[tex]\[ S_2 = \frac{2}{2} \left(2a + (2-1)d\right) = 2a + d \][/tex]
- For [tex]\(t_5\)[/tex]:
[tex]\[ t_5 = a + 4d \][/tex]
Substituting these into the given equation [tex]\(S_2 = \frac{2}{3} t_5\)[/tex]:
[tex]\[ 2a + d = \frac{2}{3} (a + 4d) \][/tex]
Solving for [tex]\(a\)[/tex] and [tex]\(d\)[/tex]:
[tex]\[ 3(2a + d) = 2(a + 4d) \][/tex]
[tex]\[ 6a + 3d = 2a + 8d \][/tex]
Simplifying:
[tex]\[ 4a = 5d \quad \Rightarrow \quad a = \frac{5}{4} d \quad \text{(Equation 1)} \][/tex]
3. Given: [tex]\( S_4 = t_{10} + 3 \)[/tex]
Let's express [tex]\(S_4\)[/tex] and [tex]\(t_{10}\)[/tex]:
- For [tex]\(S_4\)[/tex]:
[tex]\[ S_4 = \frac{4}{2} \left(2a + (4-1)d\right) = 2(2a + 3d) = 4a + 6d \][/tex]
- For [tex]\(t_{10}\)[/tex]:
[tex]\[ t_{10} = a + 9d \][/tex]
Substituting these into the given equation [tex]\(S_4 = t_{10} + 3\)[/tex]:
[tex]\[ 4a + 6d = a + 9d + 3 \][/tex]
Simplifying:
[tex]\[ 4a + 6d = a + 9d + 3 \][/tex]
[tex]\[ 4a - a + 6d - 9d = 3 \][/tex]
[tex]\[ 3a - 3d = 3 \][/tex]
[tex]\[ a - d = 1 \quad \text{(Equation 2)} \][/tex]
4. Solving Equations 1 and 2
From Equation 1: [tex]\( a = \frac{5}{4} d \)[/tex]
Substitute [tex]\(a\)[/tex] into Equation 2:
[tex]\[ \frac{5}{4} d - d = 1 \][/tex]
[tex]\[ \frac{5d - 4d}{4} = 1 \][/tex]
[tex]\[ \frac{d}{4} = 1 \][/tex]
[tex]\[ d = 4 \][/tex]
Now find [tex]\(a\)[/tex] using [tex]\( d = 4 \)[/tex] in Equation 1:
[tex]\[ a = \frac{5}{4} d = \frac{5}{4} \cdot 4 = 5 \][/tex]
So, the common difference is [tex]\( \boxed{4} \)[/tex] and the first term is [tex]\( \boxed{5} \)[/tex].
### Part (b)
Given that [tex]\( S_{p+2} - S_p = 110 \)[/tex].
1. Expressions for [tex]\( S_p \)[/tex] and [tex]\( S_{p+2} \)[/tex]
- For [tex]\( S_p \)[/tex]:
[tex]\[ S_p = \frac{p}{2} (2a + (p-1)d) \][/tex]
- For [tex]\( S_{p+2} \)[/tex]:
[tex]\[ S_{p+2} = \frac{p+2}{2} (2a + (p+2-1)d) = \frac{p+2}{2} (2a + (p+1)d) \][/tex]
2. Finding the difference [tex]\( S_{p+2} - S_p \)[/tex]
[tex]\[ S_{p+2} - S_p = \frac{p+2}{2} (2a + (p+1)d) - \frac{p}{2} (2a + (p-1)d) \][/tex]
Simplify each part:
[tex]\[ = \frac{(p+2)(2a + (p+1)d) - p(2a + (p-1)d)}{2} \][/tex]
Substitute [tex]\( a = 5 \)[/tex] and [tex]\( d = 4 \)[/tex]:
[tex]\[ = \frac{(p+2)(2(5) + (p+1)(4)) - p(2(5) + (p-1)(4))}{2} \][/tex]
[tex]\[ = \frac{(p+2)(10 + 4p + 4) - p(10 + 4p - 4)}{2} \][/tex]
[tex]\[ = \frac{(p+2)(4p + 14) - p(4p + 6)}{2} \][/tex]
[tex]\[ = \frac{(4p^2 + 14p + 8p + 28) - (4p^2 + 6p)}{2} \][/tex]
[tex]\[ = \frac{4p^2 + 22p + 28 - 4p^2 - 6p}{2} \][/tex]
[tex]\[ = \frac{16p + 28}{2} \][/tex]
[tex]\[ = 8p + 14 \][/tex]
Given: [tex]\( S_{p+2} - S_p = 110 \)[/tex]:
[tex]\[ 8p + 14 = 110 \][/tex]
[tex]\[ 8p = 96 \][/tex]
[tex]\[ p = 12 \][/tex]
So, the value of [tex]\(p\)[/tex] is [tex]\( \boxed{12} \)[/tex].
### Part (a)
We need to find the common difference [tex]\(d\)[/tex] and the first term [tex]\(a\)[/tex] of the arithmetic series.
1. Expressions for [tex]\( t_n \)[/tex] and [tex]\( S_n \)[/tex]
- The [tex]\(n\)[/tex]-th term of an arithmetic series is given by:
[tex]\[ t_n = a + (n-1)d \][/tex]
- The sum of the first [tex]\(n\)[/tex] terms [tex]\(S_n\)[/tex] of the series is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
2. Given: [tex]\( S_2 = \frac{2}{3} t_5 \)[/tex]
First, let's express [tex]\(S_2\)[/tex] and [tex]\(t_5\)[/tex]:
- For [tex]\(S_2\)[/tex]:
[tex]\[ S_2 = \frac{2}{2} \left(2a + (2-1)d\right) = 2a + d \][/tex]
- For [tex]\(t_5\)[/tex]:
[tex]\[ t_5 = a + 4d \][/tex]
Substituting these into the given equation [tex]\(S_2 = \frac{2}{3} t_5\)[/tex]:
[tex]\[ 2a + d = \frac{2}{3} (a + 4d) \][/tex]
Solving for [tex]\(a\)[/tex] and [tex]\(d\)[/tex]:
[tex]\[ 3(2a + d) = 2(a + 4d) \][/tex]
[tex]\[ 6a + 3d = 2a + 8d \][/tex]
Simplifying:
[tex]\[ 4a = 5d \quad \Rightarrow \quad a = \frac{5}{4} d \quad \text{(Equation 1)} \][/tex]
3. Given: [tex]\( S_4 = t_{10} + 3 \)[/tex]
Let's express [tex]\(S_4\)[/tex] and [tex]\(t_{10}\)[/tex]:
- For [tex]\(S_4\)[/tex]:
[tex]\[ S_4 = \frac{4}{2} \left(2a + (4-1)d\right) = 2(2a + 3d) = 4a + 6d \][/tex]
- For [tex]\(t_{10}\)[/tex]:
[tex]\[ t_{10} = a + 9d \][/tex]
Substituting these into the given equation [tex]\(S_4 = t_{10} + 3\)[/tex]:
[tex]\[ 4a + 6d = a + 9d + 3 \][/tex]
Simplifying:
[tex]\[ 4a + 6d = a + 9d + 3 \][/tex]
[tex]\[ 4a - a + 6d - 9d = 3 \][/tex]
[tex]\[ 3a - 3d = 3 \][/tex]
[tex]\[ a - d = 1 \quad \text{(Equation 2)} \][/tex]
4. Solving Equations 1 and 2
From Equation 1: [tex]\( a = \frac{5}{4} d \)[/tex]
Substitute [tex]\(a\)[/tex] into Equation 2:
[tex]\[ \frac{5}{4} d - d = 1 \][/tex]
[tex]\[ \frac{5d - 4d}{4} = 1 \][/tex]
[tex]\[ \frac{d}{4} = 1 \][/tex]
[tex]\[ d = 4 \][/tex]
Now find [tex]\(a\)[/tex] using [tex]\( d = 4 \)[/tex] in Equation 1:
[tex]\[ a = \frac{5}{4} d = \frac{5}{4} \cdot 4 = 5 \][/tex]
So, the common difference is [tex]\( \boxed{4} \)[/tex] and the first term is [tex]\( \boxed{5} \)[/tex].
### Part (b)
Given that [tex]\( S_{p+2} - S_p = 110 \)[/tex].
1. Expressions for [tex]\( S_p \)[/tex] and [tex]\( S_{p+2} \)[/tex]
- For [tex]\( S_p \)[/tex]:
[tex]\[ S_p = \frac{p}{2} (2a + (p-1)d) \][/tex]
- For [tex]\( S_{p+2} \)[/tex]:
[tex]\[ S_{p+2} = \frac{p+2}{2} (2a + (p+2-1)d) = \frac{p+2}{2} (2a + (p+1)d) \][/tex]
2. Finding the difference [tex]\( S_{p+2} - S_p \)[/tex]
[tex]\[ S_{p+2} - S_p = \frac{p+2}{2} (2a + (p+1)d) - \frac{p}{2} (2a + (p-1)d) \][/tex]
Simplify each part:
[tex]\[ = \frac{(p+2)(2a + (p+1)d) - p(2a + (p-1)d)}{2} \][/tex]
Substitute [tex]\( a = 5 \)[/tex] and [tex]\( d = 4 \)[/tex]:
[tex]\[ = \frac{(p+2)(2(5) + (p+1)(4)) - p(2(5) + (p-1)(4))}{2} \][/tex]
[tex]\[ = \frac{(p+2)(10 + 4p + 4) - p(10 + 4p - 4)}{2} \][/tex]
[tex]\[ = \frac{(p+2)(4p + 14) - p(4p + 6)}{2} \][/tex]
[tex]\[ = \frac{(4p^2 + 14p + 8p + 28) - (4p^2 + 6p)}{2} \][/tex]
[tex]\[ = \frac{4p^2 + 22p + 28 - 4p^2 - 6p}{2} \][/tex]
[tex]\[ = \frac{16p + 28}{2} \][/tex]
[tex]\[ = 8p + 14 \][/tex]
Given: [tex]\( S_{p+2} - S_p = 110 \)[/tex]:
[tex]\[ 8p + 14 = 110 \][/tex]
[tex]\[ 8p = 96 \][/tex]
[tex]\[ p = 12 \][/tex]
So, the value of [tex]\(p\)[/tex] is [tex]\( \boxed{12} \)[/tex].
