Question

If the `pth , qth` and `rth` terms of a G.P. are `a , b , c` respectively, prove that: `a^(q-r).b^(r-p)c^(p-q)=1.`

Answer 1

Let A be the first term and R be the common ratio of the given GP. Then,
`T_(p)=a rArr a=AR^((p-1))`
`T_(q)=b rArr b=AR^((q-1))`
`T_(r)=c rArr c= AR^((r-1))`.
`:. a^((q-r)).b^((r-p)).c^((p-q))`
`={AR^((p-1))}^((q-r)).{AR^((q-1))}^((r-p)).{AR^((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^(p(q-r)+q(r-p)+r(p-q)+(r-q)+(p-r)+(q-p))`
`=(1xxR^(0))=(1xx1)=1`.
Hence, `a^((q-r)).b^((r-p)).c^((p-q))=1`.