If the ratio of sum to n terms of two A.P's is (5n+7): (3n+2), then the ratio of their 17th terms is

Options

A. `172:99`
B. `172:101`
C. `175:99`
D. `175:101`

Answer

To solve the problem, we need to find the ratio of the 17th terms of two arithmetic progressions (APs) given the ratio of their sums to n terms.

### Step-by-Step Solution:

1. **Understanding the Sum of n Terms of an AP**:
The sum of the first n terms of an arithmetic progression 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.

2. **Setting Up the Given Ratio**:
We are given that the ratio of the sums of the first n terms of two APs is:
\[ \frac{S_{n1}}{S_{n2}} = \frac{5n + 7}{3n + 2} \]
Using the formula for the sum of n terms, we can write:
\[ \frac{\frac{n}{2} \left(2a_1 + (n-1)d_1\right)}{\frac{n}{2} \left(2a_2 + (n-1)d_2\right)} = \frac{5n + 7}{3n + 2} \]
The \( \frac{n}{2} \) cancels out, leading to:
\[ \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n + 7}{3n + 2} \]

3. **Finding the 17th Terms**:
The 17th term of an AP is given by:
\[ T_n = a + (n-1)d \]
Thus, the 17th terms for the two APs are:
\[ T_{1} = a_1 + 16d_1 \]
\[ T_{2} = a_2 + 16d_2 \]
We need to find the ratio:
\[ \frac{T_{1}}{T_{2}} = \frac{a_1 + 16d_1}{a_2 + 16d_2} \]

4. **Substituting n = 17 into the Ratio**:
We substitute \(n = 17\) into the earlier ratio:
\[ \frac{2a_1 + 16d_1}{2a_2 + 16d_2} = \frac{5(17) + 7}{3(17) + 2} \]
Simplifying the right side:
\[ \frac{85 + 7}{51 + 2} = \frac{92}{53} \]

5. **Setting Up the Equation**:
Now, we have:
\[ \frac{2a_1 + 16d_1}{2a_2 + 16d_2} = \frac{92}{53} \]
To find the ratio of the 17th terms:
\[ \frac{a_1 + 16d_1}{a_2 + 16d_2} = \frac{1}{2} \cdot \frac{92}{53} = \frac{92}{106} = \frac{46}{53} \]

6. **Conclusion**:
The ratio of the 17th terms of the two APs is:
\[ \frac{46}{53} \]