If the \(n\)th terms of APs \(9,7,5,\ldots\) and \(24,21,18,\ldots\) are equal, find \(n\) and that term.
\(n=16\), term = \(-21\)
Step 1: Recall the formula for the \(n\)th term of an AP:
\(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
Step 2: For the first AP \(9, 7, 5, \ldots\):
First term \(a_1 = 9\), common difference \(d = 7 - 9 = -2\).
So, \(n\)th term = \(9 + (n-1)(-2) = 9 - 2n + 2 = 11 - 2n\).
Step 3: For the second AP \(24, 21, 18, \ldots\):
First term \(a_1 = 24\), common difference \(d = 21 - 24 = -3\).
So, \(n\)th term = \(24 + (n-1)(-3) = 24 - 3n + 3 = 27 - 3n\).
Step 4: Since the two \(n\)th terms are equal, we set them equal:
\(11 - 2n = 27 - 3n\)
Step 5: Solve for \(n\):
\(11 - 2n = 27 - 3n\)
Bring terms together: \(-2n + 3n = 27 - 11\)
\(n = 16\)
Step 6: Find the term by substituting \(n=16\):
In first AP: \(11 - 2(16) = 11 - 32 = -21\)
In second AP: \(27 - 3(16) = 27 - 48 = -21\)
Final Answer: \(n = 16\) and the common term is \(-21\).