Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

If \[\log (x + z) + \log (x + z - 2y) = 2\log (x - z)\],then \[x,y,z\] are in
A. H.P.
B. G.P.
C. A.P.
D. None of these

Answer
VerifiedVerified
161.7k+ views
Hint
AP, GP, and horsepower symbolize the common or mean of the series. expected value, mean, and mean value, severally, square measure denoted by the letters AM, GM, and HM. The abbreviations AP, GP, and horsepower symbolize progression, progression, and patterned advance, severally. AN progression could be a set of integers with a set distinction between any 2 succeeding numbers.
 By considering the reciprocals of the progression that doesn't contain zero, a patterned advance (HP) is outlined as a sequence of real numbers. The AP value per invoice is calculated by dividing the whole variety of invoices paid throughout a specified amount of your time by the whole prices incurred to pay those invoices.
Formula used:
If \[x,y,z\] are in AP
\[(y - x) = (z - y)\]
If \[x,y,z\] are in GP
\[{y^2} = xz\].
If \[x,y,z\] are in HP
\[\frac{1}{x} + \frac{1}{z} = \frac{2}{y}\]
\[\log (x z) = \log (x ) + \log (z)\]
Complete step-by-step solution
The given equation is
\[2\log (x - z) - \log (x - 2y + z) = \log (x + z)\]and
\[2\log (x - z) = \log (x - 2y + z) + \log (x + z)\]
So, \[{(x - z)^2} = (x + z)((x + z) - 2y)\]
This equation can also be written as
\[2y(x + z) = {(x + z)^2} - {(x - z)^2}\]
This equation is expanded as
\[2y(x + z) = (x + z + x - z)(x + z - x + z)\]
The equation is solved as
\[ = 2y(x + z) = 4xz\]
Hence, \[\frac{1}{x} + \frac{1}{z} = \frac{2}{y}\]
So, the H.P series has \[x,y,z\]
Therefore, the correct option is A.

Note
The first n terms of the arithmetic sequence are added together to form the sum of the first n terms of AP. AP, GP, and HP stand for the average or mean of the series. In mathematics, a set of numbers is referred to as an HP if the reciprocals of the terms are in AP. In a geometric progression (GP), the common ratio is multiplied by the previous term to produce each succeeding term.