Let the A.P. be given by a, a+d, a+2d, a+3d,..... q = a + (p -1)d ... ( i ) p = a + (q -1)d ... ( ii ) Subtracting ( ii ) from ( i ), q - p = (p - q)d => d = -1 Putting d = -1 in ( i ), a = p + q - 1 nte term = a + (n - 1)d Substituting the values of a and d, nth term = p + q - 1 + (n -1)(-1) = p + q - n.That's all🙂

Proof:

Let's consider an arithmetic progression (AP) with the first term as 'a' and the common difference as 'd'.

Given that the pth term of the AP is q, we can express it as:
q = a + (p-1)d ---(1)

Similarly, the qth term of the AP is p, which can be expressed as:
p = a + (q-1)d ---(2)

Deriving an expression for 'a' in terms of p and q:

To find the value of 'a', we can subtract equation (1) from equation (2):

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

Simplifying the expression, we get:

p - q = qd - pd + d

Rearranging the terms, we have:

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

Dividing both sides by (q - p + 1), we get:

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

Now, substituting the value of 'd' back into equation (1), we can find 'a':

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

Expanding the equation, we have:

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

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

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

Simplifying further, we get:

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

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

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

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

Dividing both sides by (q - p + 1), we obtain:

q = a - (p - q)

Rearranging the terms, we have:

a = q + (p - q)

Therefore, the nth term of the AP can be expressed as:
a + (n - 1)d = (q + (p - q)) + (n - 1)(p - q)
=> a + (n - 1)d = p + (n - 1)(p - q)
=> a + (n - 1)d = p + np - nq - p + q
=> a + (n - 1)d = np - nq + q
=> a + (n - 1)d = (p - q)n + q

Therefore, the nth term of the AP is given by:
a + (n - 1)d = (p - q)n + q
which can be simplified as:
a + (n