Show that if three quantities be in any two of the three progression, a.p., g.p. And h.p., they will be also in the third progression.
To prove that if three quantities are in any two of the three progressions (arithmetic progression, geometric progression, and harmonic progression), they will also be in the third progression, we can consider each case separately.
Let's denote the three quantities as a, b, and c.
1. Assuming a and b are in arithmetic and geometric progression:
In an arithmetic progression, the difference between consecutive terms is constant. Let's assume the common difference is d.
If a and b are in arithmetic progression, we have:
a = x
b = x + d
In a geometric progression, the ratio between consecutive terms is constant. Let's assume the common ratio is r.
If a and b are in geometric progression, we have:
a = y
b = yr
From the given conditions, we can equate the values of a, b, and c:
x = y (equation 1)
x + d = yr (equation 2)
By combining equation 1 and equation 2, we can solve for y:
y = x + d/r
Therefore, c can also be expressed in a geometric progression as:
c = y*r = (x + d/r)*r = x + d
Hence, c is in an arithmetic progression as well.
2. Assuming a and b are in arithmetic and harmonic progression:
In an arithmetic progression, the difference between consecutive terms is constant. Let's assume the common difference is d.
If a and b are in arithmetic progression, we have:
a = x
b = x + d
In a harmonic progression, the reciprocals of the terms are in arithmetic progression. Let's assume the common difference is h.
If a and b are in harmonic progression, we have:
1/a = y
1/b = y + h
From the given conditions, we can equate the values of a, b, and c:
x = 1/y (equation 3)
x + d = 1/(y + h) (equation 4)
By combining equation 3 and equation 4, we can solve for y:
y = 1/(x + d/h)
Therefore, c can also be expressed in a harmonic progression as:
1/c = y + h = 1/(x + d/h) + h
Rearranging the equation, we get:
1/c = (1 + h*(x + d/h))/(x + d/h)
1/c = (1 + hx + d)/(x + d/h)
c = (x + d/h)/(1 + hx + d)
Hence, c is in a harmonic progression as well.
3. Assuming a and b are in geometric and harmonic progression:
In a geometric progression, the ratio between consecutive terms is constant. Let's assume the common ratio is r.
If a and b are in geometric progression, we have:
a = x
b = xr
In a harmonic progression, the reciprocals of the terms are in arithmetic progression. Let's assume the common difference is h.
If a and b are in harmonic progression, we have:
1/a = y
1/b = yr + h
From the given conditions, we can equate the values of a, b, and c:
x = 1/y (equation 5)
xr = 1/(yr + h) (equation 6)
By combining equation 5 and equation 6, we can solve for y:
y = 1/(x + hr/xr)
Therefore, c can also be expressed in a harmonic progression as:
1/c = y + h = 1/(x + hr/xr) + h
Rearranging the equation, we get:
1/c = (1 + h*(x + hr/xr))/(x + hr/xr)
1/c = (1 + hx + hr/x)/(x + hr/xr)
c = (x + hr/x)/(1 + hx + hr/x)
Hence, c is in a harmonic progression as well.
From all these cases, we can conclude that if three quantities are in any two of the three progressions (arithmetic progression, geometric progression, and harmonic progression), they will also be in the third progression.