If the pth term of an A. P. is q and the qth term is p. Prove that its nth term is (p+q-n).

pth term of an AP = q

qth term = p

nth term of A.P. is (p+q-n).

We know that,

nth term of an AP (an) = a + (n - 1)d

Hence,

⟹ a + (p - 1)d = q

⟹ a + pd - d = q

⟹ a = q - pd + d -- equation (1)

Similarly,

⟹ a + (q - 1)d = p

Substitute the value of a from equation (1).

⟹ q - pd + d + qd - d = p

⟹ qd - pd = p - q

⟹ - d(p - q) = p - q

⟹ - d = 1

⟹ d = - 1

Substitute the value of d in equation (1).

⟹ a = q - p( - 1) + ( - 1)

⟹ a = q + p - 1

Now,

an = q + p - 1 + (n - 1)( - 1)

⟹ an = q + p - 1 - n + 1

⟹ an = p + q - n

Hence, Proved.

I hope it will help you.

Regards.

Step-by-step explanation:

ANSWER

pth term = q

a+(p−1)d=q

qth term = p

a+(q−1)d=p

Solving these equations, we get,

d=−1

a=(p+q−1)

Thus,

nth term = a+(n−1)d=(p+q−1)+(n−1)×(−1)=(p+q−n)

the answer is 62