If the 9th term of an A. P is 0, prove that its 29th term is twice its 19th term

Proved below.

Step-by-step explanation:

a9 = a1 + 8d = 0 where a1 = first term and d = common difference.

we need to prove that

a1 + 28d = 2 (a1 + 18d

simplifying:-

a1 + 36d - 28d = 0

a1 + 8d = 0 which is what we are given.

Therefore the proposition is true.

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1) d]

⇒ a + (9-1) d = 0

⇒ a + 8d = 0 ⇒ a = - 8d ... (i)

Now, its 19th term, T19 = a + (19-1) d

= - 8d + 18d [from Eq. (i) ]

= 10d ... (ii)

and its 29th term, T29 = a + (29-1) d

= - 8d + 28d [from Eq. (i) ]

= 20d = 2 * T19

Hence, its 29th term is twice its 19th term