if pth term of a gp is p and qth term is q.prove that the nth term is (p (to the power n-q) /q (to the power n-p) ) to the power 1/p-q

Let the first term of the G.P. be *a*. The common ratio be *r*.

Then the G.P. will be

*a*, *ar*, *ar*^{2} .......... *ar ^{n}*

^{–1}

**Given:**

*p*^{th} term (*a*_{p}) = *p*

⇒*ar*^{p-1} = *p*

and *q*^{th} term (*a _{q}*) =

*q*

⇒*ar*^{q-1} = *q*

from (1) and (2), we get

Putting the value of *r* in (1), we get

Hence *n*^{th} term (*a _{n}*) =

*ar*

^{n}^{–1}

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