If the sums of n terms of two Aps are in the ratio ( 2n+3) :(3n+2) , find the ratio of their 10th terms.

Explanation

To solve the problem, we need to find the ratio of the 10th terms of two arithmetic progressions (APs) given that the sums of their first n terms are in the ratio (2n + 3) : (3n + 2).

### 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 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. **Set Up the Sums for Two APs**:
Let’s denote the first AP by \( A_1 \) and \( D_1 \), and the second AP by \( A_2 \) and \( D_2 \).
- For the first AP:
\[ S_{n1} = \frac{n}{2} \left(2A_1 + (n-1)D_1\right) \]
- For the second AP:
\[ S_{n2} = \frac{n}{2} \left(2A_2 + (n-1)D_2\right) \]

3. **Form the Ratio of the Sums**:
According to the problem, the ratio of the sums of the first n terms of the two APs is given by:
\[ \frac{S_{n1}}{S_{n2}} = \frac{2n + 3}{3n + 2} \]
Canceling \( \frac{n}{2} \) from both sides gives:
\[ \frac{2A_1 + (n-1)D_1}{2A_2 + (n-1)D_2} = \frac{2n + 3}{3n + 2} \]

4. **Cross-Multiplying**:
Cross-multiplying gives:
\[ (2A_1 + (n-1)D_1)(3n + 2) = (2A_2 + (n-1)D_2)(2n + 3) \]

5. **Finding the 10th Terms**:
The 10th term of an AP can be expressed as:
\[ T_{10} = A + 9D \]
Therefore, we need to find the ratio:
\[ \frac{T_{10,1}}{T_{10,2}} = \frac{A_1 + 9D_1}{A_2 + 9D_2} \]

6. **Substituting for n = 19**:
To find the ratio of the 10th terms, we can set \( n = 19 \) (since \( N - 1 = 18 \) gives us \( N = 19 \)):
- Substitute \( n = 19 \) into the ratio of sums:
\[ \frac{2(19) + 3}{3(19) + 2} = \frac{41}{59} \]

7. **Final Ratio of 10th Terms**:
Thus, the ratio of the 10th terms of the two APs is:
\[ \frac{A_1 + 9D_1}{A_2 + 9D_2} = \frac{41}{59} \]

### Conclusion:
The ratio of the 10th terms of the two APs is \( 41 : 59 \).