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

1. Break Down the Problem

We need to find the ratio of the 9th terms of two arithmetic progressions (APs) given the ratio of the sums of their first $n$ terms as $(7n + 1) : (4n + 27)$.

2. Relevant Concepts

1. The sum of the first $n$ terms of an AP is given by 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.

2. The $n$-th term of an AP can be formulated as: $T_n = a + (n-1)d$

3. Analysis and Detail

Let the sums of the first $n$ terms of the two APs be $S_{1n}$ and $S_{2n}$.

Assuming: $S_{1n} = \frac{n}{2} (2a_1 + (n-1)d_1) \quad \text{and} \quad S_{2n} = \frac{n}{2} (2a_2 + (n-1)d_2)$

The given ratio of the sums is: $\frac{S_{1n}}{S_{2n}} = \frac{7n + 1}{4n + 27}$

Equating the sums: $\frac{\frac{n}{2} (2a_1 + (n-1)d_1)}{\frac{n}{2} (2a_2 + (n-1)d_2)} = \frac{7n + 1}{4n + 27}$

This simplifies to: $\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n + 1}{4n + 27}$

Now we'll find the values of the 9th terms $T_{19}$ and $T_{29}$ for each AP: $T_{1,9} = a_1 + 8d_1 \quad \text{and} \quad T_{2,9} = a_2 + 8d_2$

4. Verify and Summarize

To find the ratio of the 9th terms, we need to derive the relationship using the coefficients of $n$ from the earlier ratio. Specifically, equating coefficients:

From the expression:

1. The coefficient of $n$: $d_1 = 7 \quad \text{and} \quad d_2 = 4$

2. The constant terms imply: $2a_1 - 8 = 1 \quad \Rightarrow \quad a_1 = 4.5$ $2a_2 - 8 = 27 \quad \Rightarrow \quad a_2 = 17.5$

Now we can calculate the ratio of the 9th terms: $\frac{T_{1,9}}{T_{2,9}} = \frac{a_1 + 8d_1}{a_2 + 8d_2} = \frac{4.5 + 8 \times 7}{17.5 + 8 \times 4} = \frac{4.5 + 56}{17.5 + 32} = \frac{60.5}{49.5}$

Final Answer

The ratio of the 9th terms is: $\frac{60.5}{49.5} \approx \frac{121}{99} \quad \text{or simplified to} \quad \frac{11}{9}$ Thus, the ratio of their 9th terms is $11:9$.