If the sum of first n terms of two A.P's are in the ratio `3n+8:7n+15` then the ratio of their 12th terms is

Options

A. `8:7`
B. `7:16`
C. 174:169
D. `13:47`

Answer

To solve the problem, we need to find the ratio of the 12th terms of two arithmetic progressions (APs) given that the sum of the first n terms of these APs is in the ratio \(3n + 8 : 7n + 15\).

### Step-by-Step Solution:

1. **Understanding the Sum of n Terms of an AP**:
The sum of the first n terms 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.

2. **Setting Up the Given Ratio**:
We are given that:
\[ \frac{S_n^{(1)}}{S_n^{(2)}} = \frac{3n + 8}{7n + 15} \]
where \(S_n^{(1)}\) and \(S_n^{(2)}\) are the sums of the first n terms of the two APs.

3. **Finding the 12th Term**:
The nth term of an AP can be expressed as:
\[ A_n = a + (n-1)d \]
To find the ratio of the 12th terms, we need to express \(A_{12}^{(1)}\) and \(A_{12}^{(2)}\):
\[ A_{12}^{(1)} = a_1 + 11d_1 \]
\[ A_{12}^{(2)} = a_2 + 11d_2 \]

4. **Using the Ratio of Sums**:
We can utilize the relationship between the sums and the terms. From the given ratio of sums, we can derive the ratio of the terms. The ratio of the nth term can be approximated by substituting \(n\) with \(2n - 1\):
\[ \frac{S_{2n-1}^{(1)}}{S_{2n-1}^{(2)}} = \frac{3(2n-1) + 8}{7(2n-1) + 15} \]

5. **Substituting n = 12**:
To find the ratio of the 12th terms, we substitute \(n = 12\):
\[ \frac{S_{23}^{(1)}}{S_{23}^{(2)}} = \frac{3(23) + 8}{7(23) + 15} \]
Simplifying this gives:
\[ \frac{69 + 8}{161 + 15} = \frac{77}{176} \]

6. **Reducing the Ratio**:
Now, we simplify \(\frac{77}{176}\):
\[ \frac{77}{176} = \frac{7}{16} \]

7. **Conclusion**:
Thus, the ratio of the 12th terms of the two APs is:
\[ A_{12}^{(1)} : A_{12}^{(2)} = 7 : 16 \]

### Final Answer:
The ratio of the 12th terms of the two APs is \(7 : 16\).