Answer:
Since the difference between any two consecutive terms is constant, a/b + c, b/c + a, and c/a + b are indeed in arithmetic progression (AP).
Step-by-step explanation:
We can prove that if b + c, c + a, and a + b are in harmonic progression (HP), then a/b + c, b/c + a, and c/a + b are in arithmetic progression (AP) using the following steps:
1. Define the harmonic mean:
The harmonic mean (HM) of two numbers x and y is given by:
HM(x, y) = 2xy / (x + y)
2. Express the given terms in terms of HM:
Since b + c, c + a, and a + b are in HP, their harmonic mean is equal to a constant value, say k. Therefore:
HM(b, c) = k
HM(c, a) = k
HM(a, b) = k
3. Rewrite the terms using HM:
From the definition of HM, we can rewrite the terms as:
b + c = 2bc / k
c + a = 2ca / k
a + b = 2ab / k
4. Express the desired terms in terms of HM:
We want to prove that a/b + c, b/c + a, and c/a + b are in AP. Let's rewrite them using HM:
a/b + c = (a * k) / (2bc) + 2bc / k
b/c + a = (b * k) / (2ca) + 2ca / k
c/a + b = (c * k) / (2ab) + 2ab / k
5. Simplify the expressions:
By rearranging the terms in each equation, we get:
a/b + c = (k^2 + 2abc) / (2bc)
b/c + a = (k^2 + 2abc) / (2ca)
c/a + b = (k^2 + 2abc) / (2ab)
6. Observe the common difference:
Notice that all three expressions have the same numerator, (k^2 + 2abc). This means the difference between any two consecutive terms is simply the difference in their denominators.
Therefore, we can conclude that:
(b/c + a) - (a/b + c) = (2ca - 2bc) / (2bc * 2ca) = constant
(c/a + b) - (b/c + a) = (2ab - 2ca) / (2ab * 2ca) = constant
Since the difference between any two consecutive terms is constant, a/b + c, b/c + a, and c/a + b are indeed in arithmetic progression (AP).
