If the `p^(t h)`,`q^(t h)`and `r^(t h)`terms of a GP are a, b and c, respectively. Prove that `a^(q-r)b^(r-p)c^(p-q)=1`.

Answer

N/a

Explanation

Let first term be A and common ratio be R of G.P.
`T_(p)=a`
`impliesA*R^(p-1)=a`
`T_(q)=b`
`A*R^(q-1)=b`
`T_(r)=c`
`impliesA.R^(r-1)=c`
LHS `=a^(q-r)*b^(r-p)*c^(p-q)`
`=(A*R^(p-1))^(q-r)*(A*R^(q-1))^(r-p)*(A*R^(r-1))^(p-q)`
`=A^(q-r)*R^((p-1)(q-r))*A^(r-p)*R^((q-1)(r-p))*A^(p-q)*R^((r-1)(p-q))`
`A^({q-r+r-p+p-q})*R({(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)})`
`A^(0).R({pq-pr-q+qr-qp-r+p+rp-rq-p=q})`
`1xxR^(0)=1=RHS" "`Hence Proved