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Question
If x, y, z are in A.P. and A1 is the A.M. of x and y and A2 is the A.M. of y and z, then prove that the A.M. of A1 and A2 is y.
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Solution
x, y, z are in A.P.
\[\therefore\] \[y = \frac{x + z}{2}\]
Now,
\[A_1\text { is the arithmetic mean of x and y } .\]
\[A_1 = \frac{x + y}{2} = \frac{x + \frac{x + z}{2}}{2} = \frac{3x + z}{4}\]
And,
\[A_2 \text { is the arithmetic mean of y and z } .\]
\[A_2 = \frac{y + z}{2} = \frac{\frac{x + z}{2} + z}{2} = \frac{3z + x}{4}\]
Let \[A_3\] be the arithmetic mean of \[A_1 \text { and } A_2\].
\[A_3 = \frac{A_1 + A_2}{2}\]
\[ = \frac{\frac{3x + z}{4} + \frac{3z + x}{4}}{2}\]
\[ = \frac{4x + 4z}{8}\]
\[ = \frac{x + z}{2}\]
\[ = y\]
Hence, proved.
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