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 solve the problem, we need to find the ratio of the nth terms of two arithmetic series given that the sum of their n terms is in the ratio of \( (7n + 1) : (4n + 27) \).

### Step-by-Step Solution:

1. **Define the Series**:
Let the first arithmetic series have the first term \( A_1 \) and common difference \( D_1 \).
Let the second arithmetic series have the first term \( A_2 \) and common difference \( D_2 \).

2. **Sum of n Terms**:
The sum of the first n terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \left( 2A + (n-1)D \right) \]
Therefore, the sum of the first n terms for the first series is:
\[ S_{n1} = \frac{n}{2} \left( 2A_1 + (n-1)D_1 \right) \]
And for the second series:
\[ S_{n2} = \frac{n}{2} \left( 2A_2 + (n-1)D_2 \right) \]

3. **Set Up the Ratio**:
According to the problem, the ratio of the sums of the two series is:
\[ \frac{S_{n1}}{S_{n2}} = \frac{7n + 1}{4n + 27} \]
Substituting the expressions for \( S_{n1} \) and \( S_{n2} \):
\[ \frac{\frac{n}{2} \left( 2A_1 + (n-1)D_1 \right)}{\frac{n}{2} \left( 2A_2 + (n-1)D_2 \right)} = \frac{7n + 1}{4n + 27} \]
This simplifies to:
\[ \frac{2A_1 + (n-1)D_1}{2A_2 + (n-1)D_2} = \frac{7n + 1}{4n + 27} \]

4. **Cross-Multiply**:
Cross-multiplying gives:
\[ (2A_1 + (n-1)D_1)(4n + 27) = (2A_2 + (n-1)D_2)(7n + 1) \]

5. **Expand Both Sides**:
Expanding both sides:
\[ 8nA_1 + 54A_1 + 4n(n-1)D_1 + 27(n-1)D_1 = 14nA_2 + 2A_2 + 7n(n-1)D_2 + (n-1)D_2 \]

6. **Collect Like Terms**:
Rearranging and collecting like terms will help us isolate terms involving \( n \).

7. **Find the nth Terms**:
The nth term of the first series is:
\[ T_{n1} = A_1 + (n-1)D_1 \]
The nth term of the second series is:
\[ T_{n2} = A_2 + (n-1)D_2 \]

8. **Ratio of nth Terms**:
We need to find the ratio:
\[ \frac{T_{n1}}{T_{n2}} = \frac{A_1 + (n-1)D_1}{A_2 + (n-1)D_2} \]

9. **Final Ratio**:
After substituting the values from the previous steps and simplifying, we can find the ratio of the nth terms.