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

Answer

To find the ratio of the 11th terms of two arithmetic progressions (APs) given the ratio of their sums of n terms, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
We are given that the ratio of the sum of n terms of two APs is \((7n + 1) : (4n + 27)\).

2. **Formulate the Sum of n Terms**:
The sum of n terms \(S_n\) of an AP 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.

3. **Set Up the Ratio of Sums**:
For the first AP:
\[ S_n^{(1)} = \frac{n}{2} \left(2A + (n-1)D\right) \]
For the second AP:
\[ S_n^{(2)} = \frac{n}{2} \left(2A' + (n-1)D'\right) \]
Therefore, the ratio of the sums is:
\[ \frac{S_n^{(1)}}{S_n^{(2)}} = \frac{2A + (n-1)D}{2A' + (n-1)D'} \]
This is given to equal:
\[ \frac{7n + 1}{4n + 27} \]

4. **Cross-Multiply the Ratios**:
We can set up the equation:
\[ (2A + (n-1)D)(4n + 27) = (2A' + (n-1)D')(7n + 1) \]

5. **Substitute \(n = 21\)**:
To find the ratio of the 11th terms, we substitute \(n = 21\):
\[ 2A + 20D \quad \text{and} \quad 2A' + 20D' \]
Thus, we have:
\[ \frac{2A + 20D}{2A' + 20D'} = \frac{7 \cdot 21 + 1}{4 \cdot 21 + 27} \]

6. **Calculate the Right Side**:
Calculate the right-hand side:
\[ 7 \cdot 21 + 1 = 148 \]
\[ 4 \cdot 21 + 27 = 111 \]
Therefore:
\[ \frac{2A + 20D}{2A' + 20D'} = \frac{148}{111} \]

7. **Simplify the Ratio**:
The ratio \( \frac{148}{111} \) can be simplified:
\[ \frac{148 \div 37}{111 \div 37} = \frac{4}{3} \]

8. **Conclusion**:
The ratio of the 11th terms of the two APs is:
\[ \frac{T_{11}^{(1)}}{T_{11}^{(2)}} = \frac{4}{3} \]

### Final Answer:
The ratio of the 11th terms of the two APs is \(4 : 3\).

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