The sum of n terms of two arithmetic series are in the ratio of `(7n + 1): (4n + 27)`. Find the ratio of their `n^(th)` term.

Answer

To find the ratio of the \( n^{th} \) terms of two arithmetic series given that the sum of their \( n \) terms is in the ratio \( (7n + 1):(4n + 27) \), we can follow these steps:

### Step 1: Understand the Sum of \( n \) Terms of an Arithmetic Series
The sum of the first \( n \) terms \( S_n \) of an arithmetic series 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 Equation for the Given Ratio
Let \( S_{n1} \) and \( S_{n2} \) be the sums of the first \( n \) terms of the two arithmetic series. According to the problem, we have:
\[ \frac{S_{n1}}{S_{n2}} = \frac{7n + 1}{4n + 27} \]

### Step 3: Express the Sums in Terms of \( a_1, d_1 \) and \( a_2, d_2 \)
Using the formula for the sum of \( n \) terms, we can write:
\[ S_{n1} = \frac{n}{2} \left(2a_1 + (n-1)d_1\right) \]
\[ S_{n2} = \frac{n}{2} \left(2a_2 + (n-1)d_2\right) \]
Thus, the ratio becomes:
\[ \frac{\frac{n}{2} \left(2a_1 + (n-1)d_1\right)}{\frac{n}{2} \left(2a_2 + (n-1)d_2\right)} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} \]

### Step 4: Set the Ratios Equal
Now we can set the two ratios equal:
\[ \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n + 1}{4n + 27} \]

### Step 5: Cross Multiply
Cross multiplying gives us:
\[ (2a_1 + (n-1)d_1)(4n + 27) = (2a_2 + (n-1)d_2)(7n + 1) \]

### Step 6: Substitute \( n = 1 \) to Find the Ratio of \( n^{th} \) Terms
To find the ratio of the \( n^{th} \) terms, we express the \( n^{th} \) term \( T_n \) as:
\[ T_{n1} = a_1 + (n-1)d_1 \]
\[ T_{n2} = a_2 + (n-1)d_2 \]
We can substitute \( n = 2 \) into the earlier derived equation to simplify our calculations:
\[ T_{n1} = a_1 + (2-1)d_1 = a_1 + d_1 \]
\[ T_{n2} = a_2 + (2-1)d_2 = a_2 + d_2 \]

### Step 7: Find the Ratio of \( n^{th} \) Terms
From our earlier equation, substituting \( n = 2 \):
\[ (2a_1 + d_1)(4(2) + 27) = (2a_2 + d_2)(7(2) + 1) \]
This simplifies to:
\[ (2a_1 + d_1)(8 + 27) = (2a_2 + d_2)(14 + 1) \]
\[ (2a_1 + d_1)(35) = (2a_2 + d_2)(15) \]
Now, we can find the ratio of the \( n^{th} \) terms:
\[ \frac{T_{n1}}{T_{n2}} = \frac{a_1 + d_1}{a_2 + d_2} \]

### Step 8: Final Ratio Calculation
After substituting and simplifying, we find:
\[ \frac{T_{n1}}{T_{n2}} = \frac{14n - 6}{8n + 23} \]

### Conclusion
Thus, the ratio of the \( n^{th} \) terms of the two arithmetic series is:
\[ \frac{14n - 6}{8n + 23} \]