Question 933333
If a,b,c,d are in H.P., prove that ab+bc+cd = 3ad
<pre>
Then the reciprocals {{{matrix(1,7,    1/a,  ",",  1/b, ",",  1/c, ",", 1/d)}}} are in AP.

Let 2k = the common difference of the AP
Let m = the mean of the four terms of the AP 

[Note: m is the number half-way between the middle two terms, so 1/b is k 
less than m and 1/a is 2k less than 1/b or 3k less than m.  Similarly 1/c
is k more than m and 1/d is 2k more than 1/c or 3k more than m.)

So {{{1/b=m-k}}}, {{{1/a=m-3k}}}, {{{1/c=m+k}}}, {{{1/d=m+3k}}}

and {{{b=1/(m-k)}}}, {{{a=1/(m-3k)}}}, {{{c=1/(m+k)}}}, {{{d=1/(m+3k)}}}

{{{ab+bc+cd}}}{{{""=""}}}{{{1/((m-3k)(m-k))+1/((m-k)(m+k))+1/((m+k)(m+3k)))}}}{{{""=""}}}

Get LCD:

{{{((m+3k)(m+k)+(m-3k)(m+3k)+(m-k)(m-3k))/((m-3k)(m-k)(m+k)(m+3k))}}}{{{""=""}}}

{{{(m^2+4mk+3k^2+m^2-9k^2+m^2-4mk+3k^2)/((m-3k)(m-k)(m+k)(m+3k))}}}{{{""=""}}}

{{{(3m^2-3k^2)/((m-3k)(m-k)(m+k)(m+3k))}}}{{{""=""}}}

{{{(3(m^2-k^2))/((m-3k)(m-k)(m+k)(m+3k))}}}{{{""=""}}}

{{{(3(m-k)(m+k))/((m-3k)(m-k)(m+k)(m+3k))}}}{{{""=""}}}
 
{{{(3cross((m-k)(m+k)))/((m-3k)cross((m-k)(m+k))(m+3k))}}}{{{""=""}}}

{{{3/((m-3k)(m+3k))}}}{{{""=""}}}

{{{3(1/(m-3k))(1/(m+3k))}}}{{{""=""}}}

{{{3ad}}}

Edwin</pre>