Question

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

Answer 1

Let the first term of the geometric progression be A and the common ratio be R.

pth term = AR^{p – 1} = a    ...(i)

qth term = AR^{q – 1} = b     ...(ii)

rth term = AR^{r – 1} = c    ...(iii)

Using q – r of equation (i), r – p of equation (ii), p – q power of equation (iii),

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 + p - q) R^((p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q))`

= `A^0. R^(p (q - r) - 1 (q - r) + q (r - p) - 1(r - p) + r (p - q) - 1(p - q))`

= `R^(pq - pr - q + r + qr- pq - r + p + rp - rp - p + q)`

= R⁰ 

= 1

Thus, the given result is proved.