IF the ratio between the sums of n terms of two arithmetic progressions is (7n+1): ( 4n+27), find the ratio of their 11th terms.

Explanation

To find the ratio of the 11th terms of two arithmetic progressions (APs) given the ratio of their sums of n terms, we can follow these steps:

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

### Step 2: Set up the equations for the two APs
Let:
- For the first AP, the first term be \( a \) and the common difference be \( d \).
- For the second AP, the first term be \( A \) and the common difference be \( D \).

The sums of the first n terms for both APs are:
\[ S_{n1} = \frac{n}{2} \left(2a + (n-1)d\right) \]
\[ S_{n2} = \frac{n}{2} \left(2A + (n-1)D\right) \]

### Step 3: Set up the ratio of the sums
According to the problem, the ratio of the sums of the first n terms is given as:
\[ \frac{S_{n1}}{S_{n2}} = \frac{7n + 1}{4n + 27} \]
Substituting the expressions for \( S_{n1} \) and \( S_{n2} \):
\[ \frac{\frac{n}{2} \left(2a + (n-1)d\right)}{\frac{n}{2} \left(2A + (n-1)D\right)} = \frac{7n + 1}{4n + 27} \]
The \( \frac{n}{2} \) cancels out:
\[ \frac{2a + (n-1)d}{2A + (n-1)D} = \frac{7n + 1}{4n + 27} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (2a + (n-1)d)(4n + 27) = (7n + 1)(2A + (n-1)D) \]

### Step 5: Find the ratio of the 11th terms
The 11th term \( T_{11} \) of an AP can be calculated using the formula:
\[ T_{11} = a + 10d \quad \text{(for the first AP)} \]
\[ T_{11} = A + 10D \quad \text{(for the second AP)} \]
Thus, the ratio of the 11th terms is:
\[ \frac{T_{11,1}}{T_{11,2}} = \frac{a + 10d}{A + 10D} \]

### Step 6: Substitute \( n = 11 \) into the ratio of sums
To find the specific ratio of the 11th terms, substitute \( n = 11 \) into our earlier ratio:
\[ \frac{2a + 10d}{2A + 10D} = \frac{7(11) + 1}{4(11) + 27} = \frac{77 + 1}{44 + 27} = \frac{78}{71} \]

### Step 7: Solve for the ratio of the 11th terms
Now, we can express the ratio of the 11th terms in terms of the ratio we found:
\[ \frac{a + 10d}{A + 10D} = \frac{(2a + 10d)/2}{(2A + 10D)/2} = \frac{78/71}{2} = \frac{78}{142} = \frac{39}{71} \]

### Final Answer
Thus, the ratio of the 11th terms of the two arithmetic progressions is:
\[ \frac{39}{71} \]