If \(7\) times the 7th term of an AP is equal to \(11\) times its 11th term, then its 18th term will be
7
11
18
0
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: Write the 7th term.
\(a_7 = a + (7-1)d = a + 6d\).
Step 3: Write the 11th term.
\(a_{11} = a + (11-1)d = a + 10d\).
Step 4: Use the condition given in the question.
7 times the 7th term = 11 times the 11th term.
So, \(7(a + 6d) = 11(a + 10d)\).
Step 5: Expand both sides.
\(7a + 42d = 11a + 110d\).
Step 6: Bring like terms together.
\(7a - 11a = 110d - 42d\).
\(-4a = 68d\).
Step 7: Simplify.
\(a = -17d\).
Step 8: Find the 18th term.
\(a_{18} = a + (18-1)d = a + 17d\).
Step 9: Substitute \(a = -17d\).
\(a_{18} = -17d + 17d = 0\).
Final Answer: The 18th term is 0 (Option D).