if x,y,z are in AP show that (xy)^-1,(zx)^-1,(yz)^-1 are in AP
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As given x, y, z are in AP.
Then, d ⇒ a₂ - a₁ = a₃ - a₂
And so, y - x = z - y
Divide xyz on both sides
y - x / xyz = z - y / xyz
y/xyz - x/xyz = z/xyz - y/xyz
1/xz - 1/yz = 1/xy - 1/xz
Then, by converse of the d ⇒ a₂ - a₁ = a₃ - a₂
1/xy, 1/xz, 1/yz are in AP.
Hence (xy)^-1,(zx)^-1,(yz)^-1 are in AP.
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☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
Then, d ⇒ a₂ - a₁ = a₃ - a₂
And so, y - x = z - y
Divide xyz on both sides
y - x / xyz = z - y / xyz
y/xyz - x/xyz = z/xyz - y/xyz
1/xz - 1/yz = 1/xy - 1/xz
Then, by converse of the d ⇒ a₂ - a₁ = a₃ - a₂
1/xy, 1/xz, 1/yz are in AP.
Hence (xy)^-1,(zx)^-1,(yz)^-1 are in AP.
___________________________________________________
☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
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