If the ratio of the sum of 'n' terms of two A.P's is (5n+4) : (9n+6), find the ratio of the 18th terms of these A.P.'s.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the sums of the first n terms of the two A.P.s
Let the first terms of the two A.P.s be \( A_1 \) and \( A_2 \), and their common differences be \( d_1 \) and \( d_2 \) respectively. The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
Thus, for the first A.P.:
\[ S_1 = \frac{n}{2} \left(2A_1 + (n-1)d_1\right) \]
And for the second A.P.:
\[ S_2 = \frac{n}{2} \left(2A_2 + (n-1)d_2\right) \]

### Step 2: Set up the ratio of the sums
According to the problem, the ratio of the sums of the two A.P.s is given as:
\[ \frac{S_1}{S_2} = \frac{5n + 4}{9n + 6} \]
Substituting the expressions for \( S_1 \) and \( S_2 \):
\[ \frac{\frac{n}{2} \left(2A_1 + (n-1)d_1\right)}{\frac{n}{2} \left(2A_2 + (n-1)d_2\right)} = \frac{5n + 4}{9n + 6} \]
The \( \frac{n}{2} \) cancels out:
\[ \frac{2A_1 + (n-1)d_1}{2A_2 + (n-1)d_2} = \frac{5n + 4}{9n + 6} \]

### Step 3: Substitute \( n = 18 \)
To find the ratio of the 18th terms, we will substitute \( n = 18 \):
\[ \frac{2A_1 + 17d_1}{2A_2 + 17d_2} = \frac{5(18) + 4}{9(18) + 6} \]
Calculating the right-hand side:
\[ 5(18) + 4 = 90 + 4 = 94 \]
\[ 9(18) + 6 = 162 + 6 = 168 \]
Thus, we have:
\[ \frac{2A_1 + 17d_1}{2A_2 + 17d_2} = \frac{94}{168} \]

### Step 4: Simplify the ratio
Now we simplify \( \frac{94}{168} \):
\[ \frac{94 \div 2}{168 \div 2} = \frac{47}{84} \]

### Step 5: Find the ratio of the 18th terms
The 18th term of the first A.P. is given by:
\[ T_{18,1} = A_1 + 17d_1 \]
And for the second A.P.:
\[ T_{18,2} = A_2 + 17d_2 \]
We need to find the ratio:
\[ \frac{T_{18,1}}{T_{18,2}} = \frac{A_1 + 17d_1}{A_2 + 17d_2} \]
Since we have already established the ratio of the sums, we can use it to find this ratio. We can express it in terms of the previous ratio:
\[ \frac{A_1 + 17d_1}{A_2 + 17d_2} = \frac{47}{84} \]

### Final Answer
Thus, the ratio of the 18th terms of the two A.P.s is:
\[ \frac{T_{18,1}}{T_{18,2}} = \frac{47}{84} \]