If x y z are in AP and à1 is the amof x and y and A2 is the am of y a

1.	If x y z are in AP and à1 is the amof 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» $\textbf{Concept:}$ $\text{If a, b and <klux>C</klux> are in A.P, then}$ $\boxed{\bf\,b=\dfrac{a+c}{2}}$ $\textbf{Given:}$ $\text{x, y, z are in A.P}$ $\text{$A_1$ is the <klux>ARITHMETIC</klux> 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 brainly.in/question/9603308