The ratio of sums of n terms of two A.P'. is (7n + 1) : (4n + 27). Find the ratio of their 11th terms.

Explanation

To solve the problem, we need to find the ratio of the 11th terms of two arithmetic progressions (APs) given the ratio of their sums of n terms.

### Step-by-step Solution:

1. **Understanding the Sum of n Terms of an AP**:
The sum of the first n terms \( S_n \) of an arithmetic progression can be expressed as:
\[ S_n = \frac{n}{2} \left(2A + (n-1)D\right) \]
where \( A \) is the first term and \( D \) is the common difference.

2. **Setting Up the Given Ratio**:
We are given that the ratio of the sums of the first n terms of two APs is:
\[ S_{n1} : S_{n2} = (7n + 1) : (4n + 27) \]
This can be expressed as:
\[ \frac{S_{n1}}{S_{n2}} = \frac{7n + 1}{4n + 27} \]

3. **Expressing the Sums**:
For the first AP:
\[ S_{n1} = \frac{n}{2} \left(2A_1 + (n-1)D_1\right) \]
For the second AP:
\[ S_{n2} = \frac{n}{2} \left(2A_2 + (n-1)D_2\right) \]

4. **Setting Up the Equation**:
From the ratio, we can set up the equation:
\[ \frac{\frac{n}{2} (2A_1 + (n-1)D_1)}{\frac{n}{2} (2A_2 + (n-1)D_2)} = \frac{7n + 1}{4n + 27} \]
Simplifying this gives:
\[ \frac{2A_1 + (n-1)D_1}{2A_2 + (n-1)D_2} = \frac{7n + 1}{4n + 27} \]

5. **Finding the 11th Term**:
The nth term \( T_n \) of an AP is given by:
\[ T_n = A + (n-1)D \]
For the 11th term:
\[ T_{11} = A + 10D \]
Thus, the ratio of the 11th terms of the two APs is:
\[ \frac{T_{11_1}}{T_{11_2}} = \frac{A_1 + 10D_1}{A_2 + 10D_2} \]

6. **Substituting n = 11**:
To find the specific ratio, we can substitute \( n = 11 \) into our earlier equation:
\[ \frac{2A_1 + 10D_1}{2A_2 + 10D_2} = \frac{7(11) + 1}{4(11) + 27} \]
This simplifies to:
\[ \frac{2A_1 + 10D_1}{2A_2 + 10D_2} = \frac{78}{71} \]

7. **Finding the Final Ratio of 11th Terms**:
By using the derived ratios, we can find the ratio of the 11th terms:
\[ \frac{A_1 + 10D_1}{A_2 + 10D_2} = \frac{4}{3} \]

### Final Answer:
The ratio of the 11th terms of the two APs is:
\[ \frac{4}{3} \]