If the ratio of the sum of n terms of two arithmetic progressions is `(3n+8):(7n+15)` then the ratio of their 12th terms is :

Options

A. `7:16`
B. `16:7`
C. `7:9`
D. None of these

Explanation

To solve the problem, we need to find the ratio of the 12th terms of two arithmetic progressions (APs) given the ratio of their sums of n terms. Let's break down the solution step by step.

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

### Step 2: Set up the ratio of the sums
Given that the ratio of the sums of n terms of two APs is:
\[ \frac{S_{n1}}{S_{n2}} = \frac{3n + 8}{7n + 15} \]
We can express this as:
\[ \frac{\frac{n}{2} \times (2a_1 + (n-1)d_1)}{\frac{n}{2} \times (2a_2 + (n-1)d_2)} = \frac{3n + 8}{7n + 15} \]
The \( \frac{n}{2} \) cancels out, leading to:
\[ \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n + 8}{7n + 15} \]

### Step 3: Substitute \( n = 23 \)
To find the 12th terms, we will substitute \( n = 23 \) into the equation:
\[ \frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{3(23) + 8}{7(23) + 15} \]
Calculating the right-hand side:
\[ 3(23) + 8 = 69 + 8 = 77 \]
\[ 7(23) + 15 = 161 + 15 = 176 \]
Thus, we have:
\[ \frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{77}{176} \]

### Step 4: Find the 12th terms
The 12th term of the first AP is given by:
\[ T_{12,1} = a_1 + 11d_1 \]
The 12th term of the second AP is:
\[ T_{12,2} = a_2 + 11d_2 \]
We need to find the ratio:
\[ \frac{T_{12,1}}{T_{12,2}} = \frac{a_1 + 11d_1}{a_2 + 11d_2} \]

### Step 5: Express \( a_1 + 11d_1 \) and \( a_2 + 11d_2 \) in terms of the previous ratio
From the ratio we derived earlier, we can express:
\[ \frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{77}{176} \]
This means:
\[ \frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{77/2}{176/2} = \frac{77}{176} \]

### Final Answer
Thus, the ratio of the 12th terms of the two APs is:
\[ \frac{T_{12,1}}{T_{12,2}} = \frac{77}{176} \]