(i) The sum of n terms of two arithmetic series are in the ratio of `(7n+1)/(4n+27)`. Find the ratio of their 11 th terms.
(ii) The sum of n terms of arithmetic progressions are in the ratio`(3n+8):(7n+15)`. Find the ratio of their 12 th terms.

Answer

To solve the given problems step by step, we will break down each part of the question.

### Part (i)

**Given:** The sum of n terms of two arithmetic series is in the ratio \(\frac{7n + 1}{4n + 27}\).

**To Find:** The ratio of their 11th terms.

1. **Formula for Sum of n Terms:**
The sum of the first n terms \(S_n\) of an arithmetic series can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \(a\) is the first term and \(d\) is the common difference.

2. **Let \(S_n\) for the first series be \(S_n\) and for the second series be \(S'_n\):**
\[ \frac{S_n}{S'_n} = \frac{7n + 1}{4n + 27} \]

3. **Substituting the formula for \(S_n\):**
\[ \frac{\frac{n}{2} \times (2a + (n-1)d)}{\frac{n}{2} \times (2a' + (n-1)d')} = \frac{7n + 1}{4n + 27} \]
The \(\frac{n}{2}\) cancels out:
\[ \frac{2a + (n-1)d}{2a' + (n-1)d'} = \frac{7n + 1}{4n + 27} \]

4. **Finding the 11th term:**
The 11th term \(a_{11}\) of the first series is:
\[ a_{11} = a + 10d \]
The 11th term \(a'_{11}\) of the second series is:
\[ a'_{11} = a' + 10d' \]

5. **Set up the equation for the 11th term ratio:**
We need to find:
\[ \frac{a_{11}}{a'_{11}} = \frac{a + 10d}{a' + 10d'} \]

6. **Equate coefficients:**
From the previous equation, we can compare coefficients of \(n\) and the constant terms:
- Coefficient of \(n\): \(d\) and \(d'\) must satisfy the ratio derived from the sum equation.
- Set \(n - 1 = 10\) to find \(n = 21\).

7. **Substituting \(n = 21\) into the ratio:**
\[ \frac{2a + 20d}{2a' + 20d'} = \frac{7(21) + 1}{4(21) + 27} = \frac{148}{111} \]

8. **Final ratio of 11th terms:**
Thus, the ratio of the 11th terms:
\[ \frac{a_{11}}{a'_{11}} = \frac{148}{111} \]

### Part (ii)

**Given:** The sum of n terms of two arithmetic progressions is in the ratio \((3n + 8):(7n + 15)\).

**To Find:** The ratio of their 12th terms.

1. **Using the same formula for the sum of n terms:**
\[ \frac{S_n}{S'_n} = \frac{3n + 8}{7n + 15} \]

2. **Substituting the formula for \(S_n\):**
\[ \frac{\frac{n}{2} \times (2a + (n-1)d)}{\frac{n}{2} \times (2a' + (n-1)d')} = \frac{3n + 8}{7n + 15} \]
Again, the \(\frac{n}{2}\) cancels out:
\[ \frac{2a + (n-1)d}{2a' + (n-1)d'} = \frac{3n + 8}{7n + 15} \]

3. **Finding the 12th term:**
The 12th term \(a_{12}\) of the first series is:
\[ a_{12} = a + 11d \]
The 12th term \(a'_{12}\) of the second series is:
\[ a'_{12} = a' + 11d' \]

4. **Set up the equation for the 12th term ratio:**
We need to find:
\[ \frac{a_{12}}{a'_{12}} = \frac{a + 11d}{a' + 11d'} \]

5. **Equate coefficients:**
From the previous equation, we can compare coefficients of \(n\) and the constant terms:
- Coefficient of \(n\): \(d\) and \(d'\) must satisfy the ratio derived from the sum equation.
- Set \(n - 1 = 11\) to find \(n = 23\).

6. **Substituting \(n = 23\) into the ratio:**
\[ \frac{2a + 22d}{2a' + 22d'} = \frac{3(23) + 8}{7(23) + 15} = \frac{77}{164} \]

7. **Final ratio of 12th terms:**
Thus, the ratio of the 12th terms:
\[ \frac{a_{12}}{a'_{12}} = \frac{77}{164} \]