Question

if p q r are in ap and x y z are in gp then x^q-r.y^r-p.z^p-q is equal to ?

Answer 1

answer is one . as x y z are in gp so
$y = \sqrt{xz } = > y^{2} = xz$
............ (1)
now p q r are in ap so
q-r = -d. r-p = 2d and p-q = -d
so.
$x ^{ - d} .y^{2d} . {z}^{ - d} = {x}^{ - d} {z}^{ - d} . {y}^{2d}$
so it becomes
= {y^2 / xz}^d = 1^d = 1 ...... from eqn (1)

Answer 2

Thank you for asking this question. Here is your answer:

p,q,r are in AP so 2q = p+r

and x,y,z are in gp so y² = xz

Now x^q-r y^r-p z^p-q = x^(p+r/2-r) y^r-p z^(p-(p-r)/2)

= x^(p-r/2) . y^r-p z^(p-r/2)

= x^(p-r/2) . (xz)^(r-p)/2 z^(p-r/2)

= x^(p-r/2) . x^(r-p/2) z^(r-p/2) z^(p-r/2)

= x^0 z^0

= 1

If there is any confusion please leave a comment below.