if the ratio of sum of first n terms of two ap is 7n+1 :4n+27 then find the ratio of thier 9th term

To find the ratio of the 9th terms of two arithmetic progressions (APs), we first break down the problem.

The sum of the first \( n \) terms of an arithmetic progression is given by the formula:
\[
S_n = \frac{n}{2} \times (2a + (n-1)d)
\]
where \( a \) is the first term and \( d \) is the common difference.

### Step 1: Analyze the Given Ratio
We are given that the ratio of the sum of the first \( n \) terms of two APs is:
\[
\frac{S_{n_1}}{S_{n_2}} = \frac{7n+1}{4n+27}
\]

Here, \( S_{n_1} \) and \( S_{n_2} \) are the sums of the first \( n \) terms of the first and second APs, respectively.

### Step 2: Use the Formula for the 9th Term
The nth term of an AP is given by:
\[
T_n = a + (n-1)d
\]
where \( a \) is the first term and \( d \) is the common difference.

For the 9th term, we need to find \( T_9 \) for both APs, which is:
\[
T_9 = a + 8d
\]

Now, we can work out the ratio of the 9th terms of the two APs. Let's denote the first term and common difference of the first AP as \( a_1 \) and \( d_1 \), and those of the second AP as \( a_2 \) and \( d_2 \).

### Step 3: Relating to the Sum
The sum of the first \( n \) terms of an AP can be related to its first term and common difference. Since the ratio of the sum of the first \( n \) terms of the two APs is provided as \( 7n + 1 : 4n + 27 \), we can infer that the ratios of the common differences and first terms will follow a similar relationship.

For large \( n \), the sum is dominated by the term involving \( n^2 \). Therefore, we approximate:
\[
\frac{S_{n_1}}{S_{n_2}} \approx \frac{d_1}{d_2} \approx \frac{7}{4}
\]
This implies that the ratio of the common differences \( d_1 : d_2 = 7 : 4 \).

### Step 4: Find the Ratio of 9th Terms
Using the fact that the ratio of the common differences is \( 7:4 \), we can now find the ratio of the 9th terms of the two APs:
\[
\frac{T_9^{(1)}}{T_9^{(2)}} = \frac{a_1 + 8d_1}{a_2 + 8d_2}
\]

For large \( n \), we assume the ratio of the first terms follows a similar pattern, so \( a_1 : a_2 = 7 : 4 \) as well.

Thus, the ratio of the 9th terms is:
\[
\frac{T_9^{(1)}}{T_9^{(2)}} = \frac{7 + 8 \times 7}{4 + 8 \times 4} = \frac{7 + 56}{4 + 32} = \frac{63}{36} = \frac{7}{4}
\]

### Final Answer:
The ratio of the 9th terms of the two arithmetic progressions is \( \boxed{7:4} \).

Would you like more details or explanations on any step?

Here are 5 related questions:
1. What is the formula for the nth term of an arithmetic progression?
2. How do you calculate the sum of the first n terms of an arithmetic progression?
3. Can the ratio of common differences always be inferred from the sum of terms?
4. How does the behavior of an arithmetic progression change as n increases?
5. What happens to the nth term of an AP if both the first term and common difference are doubled?

**Tip:** Always check the leading coefficients in AP sum ratios for large \( n \), as they help in determining the common differences directly.

If the ratio of sum of first n terms of two APs is 7n+1 : 4n+27, then find the ratio of their 9th term.

Discover how to find the ratio of the 9th term of two arithmetic progressions (APs) given the sum of their first n terms in the ratio 7n+1 : 4n+27. This problem covers concepts of AP, the sum formula, and relationships between sums and terms. Suitable for Grades 10-12.

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