Question

If in an GP p th ,q th and r th terms are x,y,z respectively, then show that x^q-r .y^r-p.z^p-q=1.

Answer 1

Let the First term of the Geometric Progression be : a

Let the Common Ratio of the Geometric Progression be : c

We know that : The nth term of a G.P is given by : a.cⁿ⁻¹

Given : The pth term of the G.P is x

$\implies a.c^p^-^1 = x$

$\implies x^q^-^r = (a)^q^-^r(c^p^-^1)^q^-^r$

Given : The qth term of the G.P is y

$\implies a.c^q^-^1 = y$

$\implies y^r^-^p = (a)^r^-^p(c^q^-^1)^r^-^p$

Given : The rth term of the G.P is z

$\implies a.c^r^-^1 = z$

$\implies z^p^-^q = (a)^p^-^q(c^r^-^1)^p^-^q$

$Now : (x^q^-^r).(y^r^-^p).(z^p^-^q) =$

$\implies [(a)^q^-^r(c^p^-^1)^q^-^r].[(a)^r^-^p(c^q^-^1)^r^-^p].[(a)^p^-^q(c^r^-^1)^p^-^q]$

$\implies [(a)^q^-^r^+^r^-^p^+^p^-^q].[(c^p^q^-^p^r^-^q^+^r).(c^q^r^-^q^p^-^r^+^p).(c^r^p^-^r^q^-^p^+^q)]$

$\implies [(a)^0].[(c^p^q^-^p^r^-^q^+^r^+^q^r^-^q^p^-^r^+^p^+^r^p^-^r^q^-^p^+^q)]$

$\implies (1).(c)^0 = 1.1 = 1$

$\implies (x^q^-^r).(y^r^-^p).(z^p^-^q) = 1$