Given p^th, q^th and r^thterms of a G.P are a, b and c, respectively

Here

a_p = a = ar^p-1

a_q = b = ar^q-1

a_r = c = ar^r-1

Now,

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} × b^{r – p} × c^{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} × b^{r – p} × c^{P – q} = (a^{(q – r)} × r^{(pq – q - pr +r)}) × (a^{(r - p)} × r^{(qr – r - pq + p )}) × (a^{(p - q)} × r^{(pr – p – qr + q)})

⇒ a^{q – r} × b^{r – p} × c^{P – q} = (a^{(q – r + r – p + p - q)} × r^{(pq – q – pr +r + qr - r –pq + p +pr – p –qr + q)})

⇒ a^{q – r} × b^{r – p} × c^{P – q} = (a⁰ × r⁰) = 1

∴ a^{q – r} × b^{r – p} × c^{P – q} = 1

Hence proved.