If the pth term of an arithmetic progression is q and the qth term is p, show that its (p+q)th term is 0.

To prove that the (p+q)th term of an arithmetic progression (AP) is 0, we start by recalling the formula for the nth term of an AP, which is given by:

Formula for the nth Term

The nth term of an arithmetic progression can be expressed as:

A(n) = a + (n - 1)d

where a is the first term and d is the common difference.

Given Conditions

According to the problem:

• The pth term is q: A(p) = q
• The qth term is p: A(q) = p

Expressing the Given Terms

Using the formula for the nth term, we can write:

• For the pth term: A(p) = a + (p - 1)d = q
• For the qth term: A(q) = a + (q - 1)d = p

Setting Up the Equations

From the above, we have two equations:

1. 1. a + (p - 1)d = q
2. 2. a + (q - 1)d = p

Subtracting the Equations

Now, let's subtract the first equation from the second:

(a + (q - 1)d) - (a + (p - 1)d) = p - q

This simplifies to:

((q - 1) - (p - 1))d = p - q

Which can be rewritten as:

(q - p)d = p - q

Analyzing the Result

Rearranging gives:

(q - p)d + (p - q) = 0

This implies:

(q - p)(d - 1) = 0

Finding the (p+q)th Term

Now, we need to find the (p+q)th term:

A(p + q) = a + (p + q - 1)d

Using the earlier equations, we can express a in terms of d:

From the first equation, we have:

a = q - (p - 1)d

Substituting this into the (p + q)th term gives:

A(p + q) = (q - (p - 1)d) + (p + q - 1)d

This simplifies to:

A(p + q) = q - (p - 1)d + (p + q - 1)d = q + (q - p)d

Final Step

Since we established that (q - p)d = p - q, we can substitute:

A(p + q) = q + (p - q) = 0

Thus, we conclude that the (p + q)th term of the arithmetic progression is indeed:

0