If the `p t h ,\ q t h\ a n d\ r t h` terms of `a` G.P. are `a ,\ b ,\ c` respectively, prove that: `a^((q-r))dot^(\ )b^((r-p))dotc^((p-q))=1.`

Explanation

Let A , D are the first term and common difference difference of Ap and x, R are the first term and common ratio fo Gp respectively .
According to the given condition ,
`A+(p-1)d=2`…(i)
`A+(q-1) d=b`***(iii)
` A +(r-1) d=c`...(iii)
and `a=xR^(p-1)` ...(iv)
`b=xR^(p-1)` ...(v)
`c=xR^(p-1)` ***(vi)
On subtracting Eq,(ii) from Eq (i) we get
`d(p-1-q+1)=a-b`
`a-b=d(p-q)`***(vii)
On subtracting Eq,(iii) from Eq (ii) we get
`d(p-1-r+1)=b-c`
`implies b-c=d(q-r)` ...(viii)
On subtracting Eq,(i) from Eq (iii) we get
`d(r-1-p+1)=c-a`
`c-a=d (r-p)` ....(ix)
taking LHS `=a^(b-c) b^(c-a) c^(a-b)`
using Eqs (iv) ,(v) ,(vi) and (viii) , (ix)
`LHS =(xR^(p-1))^(d(q-r) )(xR^(q-1) ) ^(d(r-p) ) (xR^(r-1))^(d(p-q))`
`=x^(d(q-1)+d(r-p)+d(p-q))R^((p-1)d(q-r)+d(r-p)+(r-1)d(p-q))`
`=x^(d(q-r+r=p-p-q))`
`R^(d(pq-pr-q+r+qr- pq -r +p rp -rp-p+q))=x^(0)r^(0)=1`
=RHS Hence proved .