Question

if x y z are in AP and à1 is the am
of x and y and A2 is the am of y and z then prove that the am of the A1 and A2 is y​

Answer 1

$\textbf{Concept:}$

$\text{If a, b and c are in A.P, then}$

$\boxed{\bf\,b=\dfrac{a+c}{2}}$

$\textbf{Given:}$

$\text{x, y, z are in A.P}$

$\text{ A_1 is the Arithmetic mean of x and y}$

$\text{ A_2 is the Arithmetic mean of y and z}$

$\textbf{To prove:}$

$\text{Arithmetic mean of A_1 and A_2 is y}$

$\textbf{Solution:}$

$\text{Since x,y,z are in A.P, we have}$

$y=\dfrac{x+z}{2}$.......(1)

$\text{Since A_1 is the arithmetic mean of x and y, we have}$

$A_1=\dfrac{x+y}{2}$.....(2)

$\text{Since A_2 is the arithmetic mean of y and z, we have}$

$A_2=\dfrac{y+z}{2}$.....(3)

$\text{Now,}$

$\text{Arithmetic mean of A_1 and A_2 }$

$=\dfrac{A_1+A_2}{2}$

$=\dfrac{(\dfrac{x+y}{2})+(\dfrac{y+z}{2})}{2}$

$=\dfrac{\dfrac{x+2y+z}{2}}{2}$

$=\dfrac{\dfrac{(x+z)+2y}{2}}{2}$

$\text{Using (1), we get}$

$=\dfrac{\dfrac{2y+2y}{2}}{2}$

$=\dfrac{\dfrac{4y}{2}}{2}$

$=\dfrac{2y}{2}$

$=y$

$\therefore\textbf{Arithmetic mean of \bf\,A_1 and \bf\,A_2 is y}$

Find more:

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation

(a) x² -18x -16 = 0

(b) x² -18x +16 = 0

(c) x² +18x -16 = 0

(d) x² +18x +16 = 0

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