If for a G.P., p^th, q^th and r^th terms are a, b and c respectively; prove that :

(q - r)log a + (r - p)log b + (p - q)log c = 0

Let first term of G.P. be A and common ratio be R.

p^th term = a

AR^{p - 1} = a.

q^th term = b

AR^{q - 1} = b.

r^th term = c

AR^{r - 1} = c.

$\Rightarrow a^{q - r}.b^{r - p}.c^{p - q} = (AR^{p - 1})^{q - r}.(AR^{q - 1})^{r - p}.(AR^{r - 1})^{p - q} \\[1em] = A^{q - r}.R^{(p - 1)(q - r)}.A^{r - p}.R^{(q - 1)(r - p)}.A^{p - q}.R^{(r - 1)(p - q)} \\[1em] = A^{q - r + r - p + p - q}.R^{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)} \\[1em] = A^0.R^{(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)} \\[1em] = A^0.R^0 \\[1em] = 1.$

Hence, proved that

$\Rightarrow a^{q - r}.b^{r - p}.c^{p - q} = 1 \\[1em] \text{Taking log on both sides} \\[1em] \Rightarrow log(a^{q - r}.b^{r - p}.c^{p - q}) = log \space 1 \\[1em] \Rightarrow log(a^{q - r}) + log(b^{r - p}) + log(c^{p - q}) = 0 \\[1em] \Rightarrow (q - r)log \space a + (r - p)log \space b + (p - q)log \space c = 0.$

Hence, proved that (q - r)log a + (r - p)log b + (p - q)log c = 0.