
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
162k+ 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.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
