The sum of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms.

Explanation

To solve the problem, we need to find the ratio of the 18th terms of two arithmetic progressions (APs) given that the sum of their first n terms is in the ratio \(5n + 4 : 9n + 6\).

### 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 Ratios:**
For the first AP, let the first term be \(a_1\) and the common difference be \(d_1\). Thus, the sum of the first n terms is:
\[ S_{n1} = \frac{n}{2} \left(2a_1 + (n-1)d_1\right) \]
For the second AP, let the first term be \(a_2\) and the common difference be \(d_2\). Thus, the sum of the first n terms is:
\[ S_{n2} = \frac{n}{2} \left(2a_2 + (n-1)d_2\right) \]
Given that the ratio of these sums is:
\[ \frac{S_{n1}}{S_{n2}} = \frac{5n + 4}{9n + 6} \]

3. **Eliminating the Common Factor:**
Since \(\frac{n}{2}\) is common in both sums, we can cancel it out:
\[ \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n + 4}{9n + 6} \]

4. **Finding the 18th Terms:**
The 18th term of the first AP is given by:
\[ T_{18,1} = a_1 + 17d_1 \]
The 18th term of the second AP is given by:
\[ T_{18,2} = a_2 + 17d_2 \]
We need to find the ratio \(\frac{T_{18,1}}{T_{18,2}}\).

5. **Using the Ratio of Sums:**
To find the ratio of the 18th terms, we can substitute \(n = 18\) into the equation we derived:
\[ \frac{2a_1 + 17d_1}{2a_2 + 17d_2} = \frac{5(18) + 4}{9(18) + 6} \]
Calculating the right-hand side:
\[ = \frac{90 + 4}{81 + 6} = \frac{94}{87} \]

6. **Finding the Ratio of 18th Terms:**
Now, we can express the ratio of the 18th terms using the values obtained:
\[ \frac{T_{18,1}}{T_{18,2}} = \frac{2a_1 + 17d_1}{2a_2 + 17d_2} = \frac{94}{87} \]

### Final Answer:
The ratio of the 18th terms of the two arithmetic progressions is:
\[ \frac{T_{18,1}}{T_{18,2}} = \frac{94}{87} \]