The ratio between the sum of n terms of two A.P.'s is 3n + 8 : 7n + 15. Then the ratio between their `12^(th)` terms is

Options

A. `5 : 7`
B. `7 : 16`
C. `12 : 11`
D. none

Answer

To solve the problem, we need to find the ratio between the 12th terms of two arithmetic progressions (A.P.s) given the ratio of their sums of n terms.

### Step-by-Step Solution:

1. **Understanding the Sum of n Terms of an A.P.**:
The sum of the first n terms (S_n) of an A.P. 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**:
According to the problem, the ratio of the sums of the first n terms of two A.P.s is given as:
\[ \frac{S_{n1}}{S_{n2}} = \frac{3n + 8}{7n + 15} \]

3. **Expressing the Sums**:
For the first A.P.:
\[ S_{n1} = \frac{n}{2} \left(2a_1 + (n - 1)d_1\right) \]
For the second A.P.:
\[ S_{n2} = \frac{n}{2} \left(2a_2 + (n - 1)d_2\right) \]

4. **Cancelling Common Terms**:
The \( \frac{n}{2} \) cancels out from both sides:
\[ \frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{3n + 8}{7n + 15} \]

5. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ (2a_1 + (n - 1)d_1)(7n + 15) = (2a_2 + (n - 1)d_2)(3n + 8) \]

6. **Comparing Coefficients**:
To find the values of \(d_1\) and \(d_2\), we can compare coefficients of \(n\) from both sides.
From the ratio, we can deduce:
- The coefficient of \(n\) gives us \(d_1 = 3\) and \(d_2 = 7\).
- The constant terms give us \(2a_1 - d_1 = 8\) and \(2a_2 - d_2 = 15\).

7. **Solving for \(a_1\) and \(a_2\)**:
- From \(2a_1 - 3 = 8\):
\[ 2a_1 = 11 \implies a_1 = \frac{11}{2} \]
- From \(2a_2 - 7 = 15\):
\[ 2a_2 = 22 \implies a_2 = 11 \]

8. **Finding the 12th Terms**:
The 12th term of an A.P. is given by:
\[ T_n = a + (n - 1)d \]
Therefore, for the first A.P.:
\[ T_{12,1} = a_1 + 11d_1 = \frac{11}{2} + 11 \times 3 = \frac{11}{2} + 33 = \frac{11 + 66}{2} = \frac{77}{2} \]
For the second A.P.:
\[ T_{12,2} = a_2 + 11d_2 = 11 + 11 \times 7 = 11 + 77 = 88 \]

9. **Finding the Ratio of the 12th Terms**:
Now, we find the ratio:
\[ \frac{T_{12,1}}{T_{12,2}} = \frac{\frac{77}{2}}{88} = \frac{77}{2 \times 88} = \frac{77}{176} = \frac{7}{16} \]

### Final Answer:
The ratio between the 12th terms of the two A.P.s is:
\[ \frac{7}{16} \]