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 need to find the ratio of the 18th terms of two arithmetic progressions (APs) given the ratio of their sums of 'n' terms. Let's break it down step by step.

### Step 1: Understand the formula for the sum of n terms of an AP
The sum of the first n terms (S_n) of an AP can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up the ratio of the sums of the two APs
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 ratio of the sums of the first n terms of these two APs is given as:
\[ \frac{S_{n1}}{S_{n2}} = \frac{5n + 4}{9n + 6} \]

Substituting the formula for the sum of n terms:
\[ \frac{\frac{n}{2} (2a_1 + (n-1)d_1)}{\frac{n}{2} (2a_2 + (n-1)d_2)} = \frac{5n + 4}{9n + 6} \]
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: Cross-multiply to simplify
Cross-multiplying gives us:
\[ (2a_1 + (n-1)d_1)(9n + 6) = (2a_2 + (n-1)d_2)(5n + 4) \]

### Step 4: Find the ratio of the 18th terms
The 18th term of an AP is given by:
\[ A_n = a + (n-1)d \]
Thus, the 18th term of the first AP is:
\[ A_{18,1} = a_1 + 17d_1 \]
And for the second AP:
\[ A_{18,2} = a_2 + 17d_2 \]
We need to find the ratio:
\[ \frac{A_{18,1}}{A_{18,2}} = \frac{a_1 + 17d_1}{a_2 + 17d_2} \]

### Step 5: Substitute \( n = 35 \)
To find the ratio, we can substitute \( n = 35 \) into our earlier equation:
\[ 2a_1 + 34d_1 = \frac{5(35) + 4}{9(35) + 6} (2a_2 + 34d_2) \]
Calculating the right-hand side:
\[ \frac{175 + 4}{315 + 6} = \frac{179}{321} \]
Thus, we have:
\[ 2a_1 + 34d_1 = \frac{179}{321} (2a_2 + 34d_2) \]

### Step 6: Rearranging for the ratio of 18th terms
From the equation:
\[ \frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{179}{321} \]
We can express it as:
\[ \frac{2(a_1 + 17d_1)}{2(a_2 + 17d_2)} = \frac{179}{321} \]
Thus, cancelling the 2's gives:
\[ \frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{179}{321} \]

### Final Answer
The ratio of the 18th terms of the two APs is:
\[ \frac{A_{18,1}}{A_{18,2}} = \frac{179}{321} \]