If the sum of first n terms of two A.P.'s are in the ratio 3n+8 : 7n+15, then the ratio of 12th term is

Options

A. `8:7`
B. `7:16`
C. `74:169`
D. `13:47`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Sum of the First n Terms of an A.P.
The sum of the first n terms \( S_n \) of an arithmetic progression (A.P.) 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 2: Set Up the Ratio of the Sums
Given that the sums of the first n terms of two A.P.s are in the ratio \( \frac{3n + 8}{7n + 15} \), we can write:
\[ \frac{S_{n1}}{S_{n2}} = \frac{3n + 8}{7n + 15} \]
Substituting the formula for \( S_n \):
\[ \frac{\frac{n}{2} \left( 2a_1 + (n - 1)d_1 \right)}{\frac{n}{2} \left( 2a_2 + (n - 1)d_2 \right)} = \frac{3n + 8}{7n + 15} \]
The \( \frac{n}{2} \) cancels out:
\[ \frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{3n + 8}{7n + 15} \]

### Step 3: Cross Multiply
Cross multiplying gives us:
\[ (2a_1 + (n - 1)d_1)(7n + 15) = (2a_2 + (n - 1)d_2)(3n + 8) \]

### Step 4: Find the 12th Term of Each A.P.
The 12th term of an A.P. is given by:
\[ T_{12} = a + (12 - 1)d = a + 11d \]
Thus, for the two A.P.s, we have:
- 12th term of A.P. 1: \( T_{12,1} = a_1 + 11d_1 \)
- 12th term of A.P. 2: \( T_{12,2} = a_2 + 11d_2 \)

### Step 5: Find the Ratio of the 12th Terms
We need to find the ratio:
\[ \frac{T_{12,1}}{T_{12,2}} = \frac{a_1 + 11d_1}{a_2 + 11d_2} \]

### Step 6: Substitute \( n = 23 \)
To find the ratio of the 12th terms, we can set \( n = 23 \) (since \( n - 1 = 22 \) leads to \( n = 23 \)):
\[ \frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{3(23) + 8}{7(23) + 15} \]
Calculating the right side:
\[ \frac{69 + 8}{161 + 15} = \frac{77}{176} \]

### Step 7: Simplify the Ratio
Now we simplify:
\[ \frac{77}{176} = \frac{7}{16} \]
Thus, the ratio of the 12th terms is:
\[ \frac{T_{12,1}}{T_{12,2}} = \frac{7}{16} \]

### Final Answer
The ratio of the 12th terms of the two A.P.s is \( \frac{7}{16} \).

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