Question

If the pth term of an AP is q and the qth term of is p prove that it's nth term is (p+q-n)

Answer 1

let 'a' be the first term

'd' is the common difference

pth term= q

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

qth term = p

a+(q-1) d =p. ----(2)

Subtract equation 2 from 1

d(p-q) = q-p

-d(q-p) = q-p

-d= 1

d= -1

substitute the value in equation 1

a= p+q-1

now

nth term= a+(n-1) d

put the value of a and d

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

nth term= p+q-1+1-n

nth term= p+q-n

##hence proved##

Answer 2

Answer:

p+q-n

Step-by-step explanation:

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)