In a given AP, if the pth term is q, and the qth term is p show that the nth term is (p+q-n).?

Let the A.P. be given by
a, a+d, a+2d, a+3d,.....
q = a + (p -1)d ... ( 1 )
p = a + (q -1)d ... ( 2 )

Subtracting ( 2 ) from ( 1 ),
q - p = (p - q)d => d = -1
Putting d = -1 in ( 1 ), 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.

let a= first term, d= common difference of AP
pth term=a+(p-1)d=q............(1)
qh term=a+(q-1)d=p..............(2)
nth term =a+(n-1)d..................(3)
(1)-(2) gives;
(p-q)d=q-p
d=-1put it in (1)
a-(p-1)=q;
a-p+1=q
a=p+q-1
putting values of a and d in (3);
nth term=p+q-1-(n-1)
=p+q-n. QED

The general formula for the nth term in an arithmetic progression with initial (zeroth) term a and difference d is

f(n) = a + n d
Substitution: f(p)=q and f(q)=p results in:
f(p) = a + p d or q = a + p d
and
f(q) = a + q d or p = a + q d

Subtraction to eliminate a and solve for d yields
q-p = (p-q) d or
d = (q-p)/(p-q) = -1

Comparison to eliminate d and solve for a yields
(q-a)/p = (p-a)/q
or
q(q-a) = p(p-a)
qq-pp = a(q-p)
or (q-p)(q+p) = a(q-p) or a=p+q

Substitution of the values for a and d into f(n) yields
f(n) = p+q - n

In AP , An=A1+(n-1)*d
Assume Ap=q, Aq=p;
hence
p=q+(q-p)*d
from that we can know d=-1

An=q+(n-p)*(-1)=q+p-n

Tn = a+(n-1)d
Now,
q=a+(p-1)d --> a-d=q-dp **
p=a+(q-1)d --> a-d=p-dq

thus, q-dp=p-dq
q-p=d(p-q)
d=-1
Put into **
a=q-(-1)(p) + (-1)=q+p-1

thus, Tn= q+p-1 + (n-1)(-1)
= q+p-1-n+1
= p+q-n QED