If \(a_9=0\), prove \(a_{29}=2a_{19}\).
True.
Step 1: Recall the formula for the \(n\)-th term of an Arithmetic Progression (AP):
\(a_n = a + (n-1)d\), where:
Step 2: For the 9th term, put \(n=9\):
\(a_9 = a + (9-1)d = a + 8d\).
Step 3: We are given that \(a_9 = 0\). So,
\(a + 8d = 0 \; \Rightarrow \; a = -8d\).
Step 4: Now, find the 29th term:
\(a_{29} = a + (29-1)d = a + 28d\).
Substitute \(a = -8d\):
\(a_{29} = -8d + 28d = 20d\).
Step 5: Next, find the 19th term:
\(a_{19} = a + (19-1)d = a + 18d\).
Substitute \(a = -8d\):
\(a_{19} = -8d + 18d = 10d\).
Step 6: Compare the results:
\(a_{29} = 20d\) and \(a_{19} = 10d\).
So, \(a_{29} = 2 \times 10d = 2a_{19}\).
Conclusion: Hence, it is proved that \(a_{29} = 2a_{19}\).