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Three positive integers x,yx, y and zz are in arithmetic progression. If y−x>2y − x > 2 and xyz=5(x+y+z)xyz = 5(x + y + z), then z−xz − x equals

Solution

✅ Correct Option: 1

Explanation:- Here x, y, 77 are in AP

⇒y=x+z2⇒x+z=2y\Rightarrow y=\frac{x+z}{2} \quad \Rightarrow x+z=2 y

Now xyz=5(x+y+z)\quad x y z=5(x+y+z)

⇒xyz=5×(2y+y)⇒xyz=5×3y⇒xz=15\begin{array}{ll} \Rightarrow & x y z=5 \times(2 y+y) \\ \Rightarrow & x y z=5 \times 3 y \\ \Rightarrow & x z=15 \end{array}

Now (x,z)(x, z) can be (3,5)&(1,15)(3,5) \&(1,15)

But if x&zx \& z are 3&53 \& 5 then y−xy-x is not greater than 2⇒x=1,z=152 \Rightarrow x=1, z=15 and y=8y=8

⇒z−x=14\Rightarrow z-x=14

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