Problem: If the 9th term of a sequence is zero, prove that the 29th term is double the 19th term.

Solution:

Given: 9th term of the sequence is zero.

We can use the formula for the nth term of an arithmetic sequence to solve this problem. The formula is:

an = a1 + (n-1)d

Where an is the nth term, a1 is the first term, n is the number of terms, and d is the common difference between the terms.

Step 1: Find the common difference

We know that the 9th term is zero. Let's substitute this into the formula to find the common difference:

0 = a1 + (9-1)d
0 = a1 + 8d

We can rearrange this equation to solve for d:

d = -a1/8

Step 2: Find the 19th term

Now that we know the common difference, we can use the formula to find the 19th term:

a19 = a1 + (19-1)d
a19 = a1 + 18d
a19 = a1 - 9a1/4
a19 = 3a1/4

Step 3: Find the 29th term

Using the same formula, we can find the 29th term:

a29 = a1 + (29-1)d
a29 = a1 + 28d
a29 = a1 - 7a1/2
a29 = -a1/2

Step 4: Prove that a29 = 2a19

We can now substitute our expressions for a19 and a29 into the equation:

a29 = 2a19
-a1/2 = 2(3a1/4)
-a1/2 = 3a1/2
-a1 = 3a1
a1 = 0

Since a1 = 0, we can conclude that the 29th term is indeed double the 19th term:

a19 = 3a1/4 = 0
a29 = -a1/2 = 0

Therefore, a29 = 2a19.

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 .