Question

Q 2

Answer 1

Let  `"a"^(1/"x")="b"^(1/"y")="c"^(1/"z")="k"`
`=> "a"^(1/"x")="k",  "b"^(1/"y")="k"` and `"c"^(1/"z") = "k"`

⇒ a = k^x, b = k^y and c = k^z

Given that a, b, c are in G.P.
⇒ b² = ac

⇒ (k^y)² = k^x x k^z

⇒ k^2y = k^x+z

⇒ 2y = x + z
⇒ x, y, z are in A.P.