For what value of \( n \), are the n-th terms of two APs 63, 65, 67, ... and 3, 10, 17, ... equal?
13
Step 1: Identify first term and common difference of each AP.
First AP: \(63, 65, 67, \ldots\)
• First term: \(a_1 = 63\)
• Common difference: \(d_1 = 65 - 63 = 2\)
Second AP: \(3, 10, 17, \ldots\)
• First term: \(a_2 = 3\)
• Common difference: \(d_2 = 10 - 3 = 7\)
Step 2: Write the general n-th term of each AP.
For an AP, \(a_n = a + (n - 1)d\).
First AP (\(T_n^{(1)}\)):
\[T_n^{(1)} = 63 + (n - 1) \cdot 2\]
\[T_n^{(1)} = 63 + 2n - 2 = 61 + 2n\]
Second AP (\(T_n^{(2)}\)):
\[T_n^{(2)} = 3 + (n - 1) \cdot 7\]
\[T_n^{(2)} = 3 + 7n - 7 = 7n - 4\]
Step 3: Use the condition that the n-th terms are equal.
We are told the n-th terms of the two APs are equal, so:
\[T_n^{(1)} = T_n^{(2)}\]
\[61 + 2n = 7n - 4\]
Step 4: Solve this equation for \(n\).
Bring all terms to one side:
\[61 + 2n = 7n - 4\]
\[61 + 2n - 7n + 4 = 0\]
\[65 - 5n = 0\]
\[5n = 65\]
\[n = \dfrac{65}{5} = 13\]
Conclusion: The n-th terms of the two APs are equal when \(n = 13\). So the required value of \(n\) is 13.