If a,b,c, are in H.P. then prove that a/b+c,b/c+a,c/a+b are also in H.P.
To prove that a/b+c, b/c+a, and c/a+b are in H.P. when a, b, c are in H.P. (Harmonic Progression), we need to show that the reciprocals of these terms are in A.P. (Arithmetic Progression).
Let's start by expanding the terms a/b+c, b/c+a, and c/a+b:
a/b+c = a/b + c/b
b/c+a = b/c + a/c
c/a+b = c/a + b/a
Now, let's take the reciprocals of these quantities:
1/(a/b+c) = 1/(a/b + c/b) = b/(a + c)
1/(b/c+a) = 1/(b/c + a/c) = c/(b + a)
1/(c/a+b) = 1/(c/a + b/a) = a/(c + b)
We can see that the reciprocals b/(a + c), c/(b + a), and a/(c + b) form an A.P. Since the reciprocals are in A.P., it implies that the original terms a/b+c, b/c+a, and c/a+b are in H.P.
Therefore, we have proved that if a, b, and c are in H.P., then a/b+c, b/c+a, and c/a+b will also be in H.P.