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Mathematics
If 2(y - a) is the harmonic mean between y - x and y - z , then x-a, y-a and z-a are in
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Q. If $2\left(y - a\right)$ is the harmonic mean between $y - x$ and $y - z$ , then $x-a, \, y-a$ and $z-a$ are in
NTA Abhyas
NTA Abhyas 2022
Sequences and Series
A
arithmetic progression
B
geometric progression
C
harmonic progression
D
none of these
Solution:
It is given that, $y-x, \, 2\left(y - a\right) \, and \, \left(y - z\right)$ are in harmonic progression.
$\Rightarrow \, \, \, \frac{1}{y - x},\frac{1}{2 \left(y - a\right)},\frac{1}{y - z}$ are in arithmetic progression.
$\Rightarrow \, \, \, \frac{1}{2 \left(y - a\right)}-\frac{1}{y - x}=\frac{1}{y - z}-\frac{1}{2 \left(y - a\right)}$
$\Rightarrow \, \, $ $\frac{2 a - y - x}{y - x}=\frac{y + z - 2 a}{y - z}$
$\Rightarrow \, \, \frac{\left(x - a\right) + \left(y - a\right)}{\left(x - a\right) - \left(y - a\right)}=\frac{\left(y - a\right) + \left(z - a\right)}{\left(y - a\right) - \left(z - a\right)}$
$\Rightarrow $ $\frac{x - a}{y - a}=\frac{y - a}{z - a}$
Hence, $x-a,y-a \, and \, z-a$ are in geometric progression