If #a+c=2b# and #ab+cd+ad=3bc#, how do I show that #a,b,c,d# are in arithmetic progression?

1 Answer
Aug 15, 2017

Play around with the equations. Knowing how to (re-)phrase your goal mathematically can help you deduce it from the given information.

Explanation:

#a, b, c, d# are in arithmetic progression if and only if #b-a = c-b = d-c#.

Can we show this?

Given: #a+c=2b#
Move one #b# to the LHS, and move #c# to the RHS.

#color(white)(=>)a+c=2b#
#=>a-b=b-c"   "#Multiply both sides by #-1#.
#=>b-a=c-b#

Now we just need to show that #d-c# is equal to either #b-a# or #c-b#.

Given: #ab+cd+ad=3bc#

Solve for #d#.

#ab+cd+ad=3bc#

#=>d(c+a)=3bc-ab#

#=>d=(b(3c-a))/(c+a)"    "#Substitute #2b=a+c#

#=>d=(cancel b(3c-a))/(2cancelb)#

#=>2d=3c-a"    "#Subtract #2c# from both sides

#=>2d-2c=3c-a-2c#

#=>2(d-c)=c-a"    "#Subtract and add #b# to the RHS

#=>2(d-c)=c-b+b-a"    "#Substitute #b-a=c-b#

#=>2(d-c)=c-b+c-b#

#=>2(d-c)=2(c-b)"    "#Divide both sides by 2

#=>d-c=c-b#

Since #b-a = c-b = d-c#, we have #a,b,c,d# in arithmetic progression.