In an A,P if the pth term is `(1)/(q) and q^(th)` terms is `(1)/(p)`. Prove that the sum of first pq term is `(1)/(2)` (pq+1) where, `(p!=q)`

Answer

N/a

Explanation

Let first term be 'a' and common difference be 'd' of A.P.
`therefore" "T_(p)=a+(p-1)d=(1)/(q)" …(1)"`
`T_(q)=a(q-1)d=(1)/(p)" …(2)"`
Subtracting equation (2) from equation (1)
`rArr" "(p-q)d=(1)/(q)-(1)/(p)=(p-q)/(pq)" "rArr d=(1)/(pq)`
Put the value of d in equation (1),
`a+(p-1)cdot(1)/(pq)=(1)/(q)" "rArr a=(1)/(pq)`
Now, sum of pq terms
`S_(pq)=(pq)/(2)[2cdot(1)/(pq)+(pq-1)(1)/(pq)]`
`=(1)/(2)(pq+1)`