The ratio between the sum of n term of two A.P. s is `3n + 8: 7n + 15.` Then find the ratio between their 12 th term

Answer

To solve the problem step by step, we will follow the same logic as in the video transcript.

### Step 1: Understand the problem
We are given the ratio of the sums of the first n terms of two arithmetic progressions (APs) as \( \frac{3n + 8}{7n + 15} \). We need to find the ratio of their 12th terms.

### Step 2: Use 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:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 3: Set up the equation
Let:
- For the first AP: first term = \( a \), common difference = \( d \)
- For the second AP: first term = \( A \), common difference = \( D \)

The ratio of the sums of the first n terms of the two APs can be written as:
\[ \frac{S_n^{(1)}}{S_n^{(2)}} = \frac{\frac{n}{2} \left( 2a + (n-1)d \right)}{\frac{n}{2} \left( 2A + (n-1)D \right)} = \frac{2a + (n-1)d}{2A + (n-1)D} \]
Given that this ratio is equal to \( \frac{3n + 8}{7n + 15} \), we have:
\[ \frac{2a + (n-1)d}{2A + (n-1)D} = \frac{3n + 8}{7n + 15} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (2a + (n-1)d)(7n + 15) = (2A + (n-1)D)(3n + 8) \]

### Step 5: Find the 12th term of both APs
The 12th term \( T_{12} \) of the first AP is:
\[ T_{12}^{(1)} = a + (12-1)d = a + 11d \]
The 12th term \( T_{12} \) of the second AP is:
\[ T_{12}^{(2)} = A + (12-1)D = A + 11D \]

### Step 6: Find the ratio of the 12th terms
We need to find the ratio:
\[ \frac{T_{12}^{(1)}}{T_{12}^{(2)}} = \frac{a + 11d}{A + 11D} \]

### Step 7: Substitute \( n \) to find a specific ratio
To find a specific ratio, we can choose \( n = 23 \) (since \( n - 1 = 22 \) gives us \( \frac{n-1}{2} = 11 \)):
Substituting \( n = 23 \) into the original ratio of sums:
\[ \frac{3(23) + 8}{7(23) + 15} = \frac{69 + 8}{161 + 15} = \frac{77}{176} \]

### Step 8: Simplify the ratio
Now we know:
\[ \frac{2a + 22d}{2A + 22D} = \frac{77}{176} \]
This means:
\[ \frac{a + 11d}{A + 11D} = \frac{77}{176} \]

### Step 9: Simplify further
Dividing both numerator and denominator by 11 gives:
\[ \frac{7}{16} \]

### Final Answer
Thus, the ratio of the 12th terms of the two APs is:
\[ \frac{7}{16} \]