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mail2abhishek99 · May 25, 2011, 11:22 PM

If a,b,c are in Harmonic Progression Prove that 1/(b-a) + 1/(b-c) = 1/a + 1/c

jcaron2 · May 26, 2011, 07:14 AM

By the definition of harmonic progression we can write out the numbers a, b, and c:

$a=\frac{m}{1+(n-1)d}$

$b=\frac{m}{1+nd}$

$c=\frac{m}{1+(n+1)d}$

where n is an integer, m is any number, and d is any number such that -(1/d) is not an integer > 0. (None of those rules is important for this proof; just treat n, m, and d like any other variable).

If you substitute the above values (to the right of the equal signs) into the formula that you're supposed to prove, you'll see that it works out. The right side of the equation is pretty trivial and works out to

$\frac 1a + \frac 1c = \frac{2(1+nd)}{m}$

The left side is a bit more tedious, but if you work it out you'll find that it comes to the same value.

$\frac 1{b-a} \; + \; \frac 1{b-c} = ?$

If you still aren't to work it out, post back and tell us where you're getting stuck. One of us will guide you through the rest of the algebra.