Algebra

First, we can write a handy equation to solve for the constant and find the 3 terms. 

Let's let the constant added to every number be x. 

Since the three terms for a geometric series, we have the equation

$\frac{100+x}{60+x} = \frac{140+x}{100+x}$

Now, when we crossmultiply and then expand everything out, we get

$(x+100)(100+x)=(x+60)(140+x)\\ 200x+x^{2}+10000=140x+x^{2}+{8400+60x}\\ 200x+x^{2}+10000=200x+x^{2}+8400$

Now, we bring all terms to one side of the equation. We have

$200x+x^{2}+10000-x^{2}-200x-8400=0\\200x+1600-200x=0 \\ 1600=0$

However, this statement is obviously not true, meaning that x is invalid. 

This also means that there are NO solutions to this given problem. 

*Note, I may have made a mistake. Not sure. 

Thanks! :)