Question 844224
if a,b,c are in HP and ab + bc + ca = 15 then ca =
<pre>
Since a,b,c are in harmonic progression, their 
reciprocals {{{1/a}}},{{{1/b}}},{{{1/c}}} form an arithmetic sequence:

Let d be the common difference.  Then

{{{1/a+d}}}{{{""+""}}}{{{1/b}}}
{{{1/b+d}}}{{{""+""}}}{{{1/c}}}

Solve each for d

{{{d}}}{{{""=""}}}{{{1/b-1/a}}}
{{{d}}}{{{""=""}}}{{{1/c-1/b}}}

Since both equal d, they are equal to each other:

{{{1/b-1/a}}}{{{""=""}}}{{{1/c-1/b}}}

Multiply through by LCD of abc

ac - bc = ab - ac

2ac = ab + bc

And since we are given that

ab + bc + ca = 15

We can substitute 2ac for ab + bc and get

2ac + ca = 15

And since ac is the same as ca

2ca + ca = 15

3ca = 15

ca = 5

Edwin</pre>