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 first 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 \( S_n \) 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 Ratios**:
Let the first AP have first term \( a_1 \) and common difference \( d_1 \), and the second AP have first term \( a_2 \) and common difference \( d_2 \). The sums of the first n terms for both APs can be written as:
\[ S_{n1} = \frac{n}{2} \left( 2a_1 + (n-1)d_1 \right) \]
\[ S_{n2} = \frac{n}{2} \left( 2a_2 + (n-1)d_2 \right) \]
According to the problem, the ratio of these sums is given by:
\[ \frac{S_{n1}}{S_{n2}} = \frac{2n + 3}{6n + 5} \]

3. **Eliminating the Common Factor**:
Since both sums have a common factor of \( \frac{n}{2} \), we can simplify the ratio:
\[ \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{2n + 3}{6n + 5} \]

4. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ (2a_1 + (n-1)d_1)(6n + 5) = (2a_2 + (n-1)d_2)(2n + 3) \]

5. **Substituting n = 25**:
To find the ratio of the 13th terms, we can substitute \( n = 25 \):
\[ (2a_1 + 24d_1)(6 \cdot 25 + 5) = (2a_2 + 24d_2)(2 \cdot 25 + 3) \]
Simplifying the constants:
\[ (2a_1 + 24d_1)(150 + 5) = (2a_2 + 24d_2)(50 + 3) \]
\[ (2a_1 + 24d_1)(155) = (2a_2 + 24d_2)(53) \]

6. **Finding the Ratio of the 13th Terms**:
The 13th term of the first AP is given by:
\[ a_1 + 12d_1 \]
The 13th term of the second AP is:
\[ a_2 + 12d_2 \]
Therefore, the ratio of the 13th terms is:
\[ \frac{a_1 + 12d_1}{a_2 + 12d_2} \]

7. **Using the Earlier Result**:
From the earlier equation, we can express the ratio of the 13th terms in terms of the coefficients we derived:
\[ \frac{a_1 + 12d_1}{a_2 + 12d_2} = \frac{(2a_1 + 24d_1) \cdot 53}{(2a_2 + 24d_2) \cdot 155} \]
By substituting the values we derived, we can find the ratio.

8. **Final Calculation**:
After substituting and simplifying, we find that the ratio of the 13th terms is:
\[ \frac{53}{155} \]

### Conclusion:
Thus, the ratio of the 13th terms of the two arithmetic progressions is \( \frac{53}{155} \).