
If \[(y - x),2(y - a)\]and \[(y - z)\] are in H.P., then \[x - a,y - a,z - a\] are in
A. A.P.
B. G.P.
C. H.P.
D. None of these
Answer
164.4k+ views
Hint
Taking the reciprocals of the arithmetic progression that does not contain zero yields a sequence of real numbers known as a harmonic progression (HP). Any phrase in a harmonic progression is regarded as the harmonic mean of its two neighbours. Making the appropriate AP series is the first step in solving an issue involving harmonic progression. If the reciprocal of the terms is in AP, a series of numbers is referred to as a harmonic progression.
AP, GP, and HP stand for the average or mean of the series. Arithmetic Mean, Geometric Mean, and Harmonic Mean, respectively, are denoted by the letters AM, GM, and HM. AM, GM, and HM are abbreviations for Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP).
Complete step-by-step solution
The given equation is \[(y - x),2(y - a),(y - z)\]
This equation is in Harmonic progression.
Here, \[\frac{1}{{(y - x)}},\frac{1}{{2(y - a)}},\frac{1}{{(y - z)}}\]are in A.P
The equation then becomes
\[ = > \frac{1}{{2(y - a)}} - \frac{1}{{(y - x)}} = \frac{1}{{(y - z)}} - \frac{1}{{2(y - a)}}\]
By solving, it becomes
\[ = > \frac{{2a - y - z}}{{y - x}} = \frac{{y + z - 2a}}{{y - z}}\]
\[ = > \frac{{(x - a) + (y - a)}}{{(x - a) - (y - a)}} = \frac{{(y - a) + (z - a)}}{{(y - a) - (z - a)}}\]
It is simplified to become
\[ = > \frac{{(x - a)}}{{(y - a)}} = \frac{{(y - a)}}{{(z - a)}}\]
\[ = > {(y - a)^2} = (x - a)(z - a)\]
So, \[(y - x),2(y - a)\]and \[(y - z)\] are in Geometric progression.
Therefore, the correct option is B.
Note
A mathematical sequence known as a geometric progression (GP) is one in which each following phrase is generated by multiplying each preceding term by a fixed integer, or "common ratio." This progression is sometimes referred to as a pattern-following geometric sequence of numbers. A series of terms is referred to as a geometric progression if each next term is produced by multiplying each previous term by a fixed amount. (GP), whereas the common ratio is the name given to the constant value.
Taking the reciprocals of the arithmetic progression that does not contain zero yields a sequence of real numbers known as a harmonic progression (HP). Any phrase in a harmonic progression is regarded as the harmonic mean of its two neighbours. Making the appropriate AP series is the first step in solving an issue involving harmonic progression. If the reciprocal of the terms is in AP, a series of numbers is referred to as a harmonic progression.
AP, GP, and HP stand for the average or mean of the series. Arithmetic Mean, Geometric Mean, and Harmonic Mean, respectively, are denoted by the letters AM, GM, and HM. AM, GM, and HM are abbreviations for Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP).
Complete step-by-step solution
The given equation is \[(y - x),2(y - a),(y - z)\]
This equation is in Harmonic progression.
Here, \[\frac{1}{{(y - x)}},\frac{1}{{2(y - a)}},\frac{1}{{(y - z)}}\]are in A.P
The equation then becomes
\[ = > \frac{1}{{2(y - a)}} - \frac{1}{{(y - x)}} = \frac{1}{{(y - z)}} - \frac{1}{{2(y - a)}}\]
By solving, it becomes
\[ = > \frac{{2a - y - z}}{{y - x}} = \frac{{y + z - 2a}}{{y - z}}\]
\[ = > \frac{{(x - a) + (y - a)}}{{(x - a) - (y - a)}} = \frac{{(y - a) + (z - a)}}{{(y - a) - (z - a)}}\]
It is simplified to become
\[ = > \frac{{(x - a)}}{{(y - a)}} = \frac{{(y - a)}}{{(z - a)}}\]
\[ = > {(y - a)^2} = (x - a)(z - a)\]
So, \[(y - x),2(y - a)\]and \[(y - z)\] are in Geometric progression.
Therefore, the correct option is B.
Note
A mathematical sequence known as a geometric progression (GP) is one in which each following phrase is generated by multiplying each preceding term by a fixed integer, or "common ratio." This progression is sometimes referred to as a pattern-following geometric sequence of numbers. A series of terms is referred to as a geometric progression if each next term is produced by multiplying each previous term by a fixed amount. (GP), whereas the common ratio is the name given to the constant value.
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