If the `p^(t h)a n dq^(t h)` terms of a G.P. are `qa n dp` respectively, show that `(p+q)^(t h)` term is `((q^p)/(p^q))^(1/(p-q))` .

Explanation

Let the first term and common ratio of GP be and r, respectively .
Accodring to the question , pth term = q
`implies a.r^(p-1)=q`
and q th term =p
`implies ar^(q-1)=p`
On dividing Eq (i) by Eq (ii) We get
`(ar^(p-1))/(ar^(q-1))=(q)/(p)`
` implies r^(p-1-q+1)=(q)/(p)`
`implies r^(p-q)=(q)/(p)implies r((q)/(p))^((1)/(p-q))`
On substituting the value of r in Eq (i) , we get
`a((q)/(p))^((p-1)/(p-q))=qimplies a=(q)/(((q)/(p))^((p-1)/(p-q)))=q. ((p)/(q))^(^(p-1)/(p-q))`
`therefore (p+q) "th term "T_(p+q)= a.r ^(p+q-1)=q/((p)/(q))^((p-1)/(p-q)). (r) ^(p+q-1)`
`=q((p)/(q))^((p-1)/(p-q))[((q)/(p))^((1)/(p-q))]^(p+q-1)=q.((p)/(q))^((p-1)/(p-q))((q)/(p))^((p+q-1)/(p-q))`
` =q.((p)/(q))^((p-1)/(p-q))((p)/(q))^((-(p+q-1))/(p-q))=q.((p)/(q))^((p-1)/(p-q)-((p+q-1))/(p-q))`
`=q.((p)/(q))^((p-1-p-q+1)/(p-q))
=q.((p)/(q))^((p-1-p-q+1)/(p-q))=q./((p)/(q))^((-q)/(p-q))`
` a=q.((p)/(q))^((p-1)/(p-q))`
Now (p+q) th term i.e., `a_(p+q)=ar^(p+q-1)`
`=.((p)/(q))^((p-1)/(p-q)).((q)/p)^((p+q-1)/(p-q))`
` =q.q^((p=q-1-p+1)/(p-q))/(p^((p+q-1-p+1)/(p-q)))=q.(q^((q)/(p-q))/(p^((q)/(p-q))))`