Given,
a9 = 0
We know that, nth term an = a + (n – 1)d
So, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ……(i)
Now,
29th term is given by a29 = a + (29 – 1)d
⇒ a29 = a + 28d
And, a29 = (a + 8d) + 20d [using (i)]
⇒ a29 = 20d ….. (ii)
Similarly, 19th term is given by a19 = a + (19 – 1)d
⇒ a19 = a + 18d
And, a19 = (a + 8d) + 10d [using (i)]
⇒ a19 = 10d …..(iii)
On comparing (ii) and (iii), it’s clearly seen that
a29 = 2(a19)
Therefore, 29th term is double the 19th term.