Question

If a,b,c be in H.P prove that 1\b-a+1\b-c=1\a+1\c

Answer 1

Hey there!

If a, b, c are in H. P ( Harmonic progression)

Then 1/a, 1/b, 1/c are in A. P.

So,

2/b = 1/a + 1/c.

2/b = a + c /ac

2ac = ( a + c) b

b = 2ac / a + c

Now,

Given,

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

L. H. S

1/b - a + 1/b - c

Substitute value of b in the equation

1/(2ac/a+c) - a + 1/(2ac/a+c) - c

= (a+c) /2ac - a(a+c) + (a+c) / 2ac - c ( a + c)

= a+ c /( 2ac - a² - ac) + a+c/ ( 2ac - ac - c²)

= a + c / ( ac - a²) + ( a + c) / ( ac - c²).

= a + c / a ( c - a) + a + c / c ( a - c)

= [a + c / c - a] [ 1/a - 1/c]

= [a + c / c - a] [ c - a / ac]

= a + c / ca

= a/ca + c/ca

= 1/c + 1/a

= R. H. S

Hence proved!