Formula:
We know that
nth term of the AP is Tn = a+(n-1)d
Where, a = First term
d = Common difference
n = number of terms
Now,
Given that
pth term of the A.P. = Tp = q
⇛a + (p-1)d = q
⇛(p-1)d = q - a
⇛d = (q-a)/(p-1) →→→→(i)
and
qth term of the A.P. = Tq = p
⇛a + (q-1) d = p
⇛ (q-1)d = p-a
⇛ d = (p-a)/(q-1) →→→→(ii)
From Eqn(i) and Eqn(ii)
⇛(q-a)/(p-1) = (p-a)/(q-1)
On applying cross multiplication then
⇛(q-a)×(q-1) = (p-a)×(p-1)
⇛q²-q-aq+a = p²-p-ap+a
⇛q²-q-aq = p²-p-ap
⇛ap-aq = p²-p+q-q²
⇛a(p-q) = (p²-q²)-(p-q)
⇛a(p-q) = (p+q)(p-q)-(p-q)
⇛ a(p-q) = (p-q){(p+q)-1}
On cancelling (p-q) both sides then
⇛a = (p+q-1) →→→→Eqn(iii)
On Substituting the value of a in Eqn(ii) then
⇛ d = [p-(p+q-1)]/(q-1)
⇛d = (p-p-q+1)/(q-1)
⇛d = (-q+1)/(q-1)
⇛d = -(q-1)/(q-1)
⇛d = -1
We have,
a = p+q-1 and d = -1
Now, rth term of the AP
⇛ Tr = a + (r-1)d
⇛Tr = (p+q-1)+(r-1)(-1)
⇛ Tr = p+q-1+(-r+1)
⇛ Tr = p+q-1-r+1
⇛ Tr = p+q-r
Therefore, Tr = p+q-r
Answer: rth term of the given A.P. is p+q-r