The sum of n terms of two arithmetic progressions are in the ratio 2n+3:6n+5, then the ratio of their 13th terms, is

Options

A. `53:155`
B. `27:87`
C. `29:89`
D. `31:89`

Answer

To solve the problem, we need to find the ratio of the 13th terms of two arithmetic progressions (APs) given that the sum of their n terms is in the ratio \( \frac{2n + 3}{6n + 5} \).

### Step-by-Step Solution:

1. **Understanding the Sum of n Terms of an AP**:
The sum of the first n terms 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 Equation**:
Let \( S_n \) be the sum of the first n terms of the first AP and \( H_n \) be the sum of the first n terms of the second AP. According to the problem, we have:
\[ \frac{S_n}{H_n} = \frac{2n + 3}{6n + 5} \]

3. **Expressing the Sums**:
For the first AP:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
For the second AP:
\[ H_n = \frac{n}{2} \left(2A + (n-1)D\right) \]

4. **Substituting into the Ratio**:
Substituting \( S_n \) and \( H_n \) into the ratio gives:
\[ \frac{\frac{n}{2} \left(2a + (n-1)d\right)}{\frac{n}{2} \left(2A + (n-1)D\right)} = \frac{2n + 3}{6n + 5} \]
The \( \frac{n}{2} \) cancels out:
\[ \frac{2a + (n-1)d}{2A + (n-1)D} = \frac{2n + 3}{6n + 5} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ (2a + (n-1)d)(6n + 5) = (2A + (n-1)D)(2n + 3) \]

6. **Finding the Ratio of the 13th Terms**:
The 13th term of the first AP is:
\[ a_{13} = a + 12d \]
The 13th term of the second AP is:
\[ A_{13} = A + 12D \]
Thus, we need to find:
\[ \frac{a + 12d}{A + 12D} \]

7. **Finding n**:
To find \( n \) such that \( \frac{n-1}{2} = 12 \), we solve:
\[ n - 1 = 24 \quad \Rightarrow \quad n = 25 \]

8. **Substituting n into the Ratio**:
Now substituting \( n = 25 \) into the ratio:
\[ \frac{2a + 24d}{2A + 24D} = \frac{2(25) + 3}{6(25) + 5} = \frac{53}{155} \]

9. **Final Ratio of the 13th Terms**:
Thus, the ratio of the 13th terms is:
\[ \frac{a + 12d}{A + 12D} = \frac{53}{155} \]

### Conclusion:
The ratio of the 13th terms of the two arithmetic progressions is \( 53:155 \).