If the Pth term of an AP is Q and Qth term of an AP is p, prove that it's Nth term is (p+ q-n).?

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Introduction:

We are given that the pth term of an AP is Q and Qth term of an AP is p. We need to prove that the Nth term of the AP is (p + q - n).

Proof:

Let us consider the AP whose pth term is Q and Qth term is p. Let the common difference of the AP be d.

Deriving Q:

The pth term of the AP can be expressed as:

Q = a + (p - 1) d

where a is the first term of the AP.

Deriving p:

Similarly, the Qth term of the AP can be expressed as:

p = a + (Q - 1) d

Deriving the Nth term:

Let us now find the Nth term of the AP. We can express the Nth term as:

T_N = a + (N - 1) d

Substituting values:

We can express a in terms of Q and d by substituting the value of Q in the equation for p:

p = a + (Q - 1) d

a = p - (Q - 1) d

Final derivation:

Substituting the value of a in the expression for the Nth term, we get:

T_N = (p - (Q - 1) d) + (N - 1) d

T_N = p + (N - 1) d - Qd

T_N = (p + Q - N) d

T_N = p + Q - N (since d is the common difference)

Therefore, we have proved that the Nth term of the AP is (p + q - n).

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