If a, b, c are in GP and a1/x = b1/y = c1/z then prove that x, y, z ar

1.	If a, b, c are in GP and a1/x = b1/y = c1/z then prove that x, y, z are in AP.
Answer» It is given that: a ^1/x = b^1/y = c^1/z Let a^1/x = b^1/y = c^1/z = k ⇒ a ^1/x = k ⇒ (a^1/x) ^x = k^x…(Taking power of x on both sides.) ⇒ a ^1/x ^{× x} = k^x ⇒ a = k^x Similarly b = k^y And c = k^z It is given that a,b,c are in G.P. ⇒ b² = ac Substituting values of a,b,c calculated above, we get: ⇒ (ky )² = k^xk ^z ⇒ k²y = k^{x + z} Comparing the powers we get, 2y = x + z Which is the required condition for x,y,z to be in A.P. Hence, proved that x,y,z, are in A.P.

If a, b, c are in GP and a1/x = b1/y = c1/z then prove that x, y, z are in AP.

Answer»

It is given that:

a ^1/x = b^1/y = c^1/z

Let a^1/x = b^1/y = c^1/z = k

⇒ a ^1/x = k

⇒ (a^1/x) ^x = k^x…(Taking power of x on both sides.)

⇒ a ^1/x ^{× x} = k^x

⇒ a = k^x

Similarly b = k^y

And c = k^z

It is given that a,b,c are in G.P.

⇒ b² = ac

Substituting values of a,b,c calculated above, we get:

⇒ (ky )² = k^xk ^z

⇒ k²y = k^{x + z}

Comparing the powers we get,

2y = x + z

Which is the required condition for x,y,z to be in A.P.

Hence, proved that x,y,z, are in A.P.