Question

If pth term of an AP is q and qth term of an AP is p, then show that its nth term is (p+q-n)

Answer 1

We are given that,
a+(p−1)d=q        −−−−−−−−−−(1)
a+(q−1)d=p        −−−−−−−−−−(2)

Subtracting (1) from (2), we get,

(p−q)=a+(q−1)d−[a+(p−1)d]
(p−q)=a+(q−1)d−a−(p−1)d
(p−q)=(q−1)d−(p−1)d
(p−q)=(q−1−p+1)d
(p−q)=(q−p)d
(−q+p)=(q−p)d
−(q−p)=(q−p)d
−1=d

From (2) we get,
a=p−(q−1)d
a=p−(q−1)(−1)    {Because d=−1}
a=p+q−1

Now we have both first term (a) and common difference (d), we can solve for nth term,

an=a+(n−1)d
i.e. an=p+q−1+(n−1)(−1)
      an=p+q−1−n+1
    
an=(p+q−n)

Hence, proved.

Answer 2

Answer:

this is the correct answer