+760 When the same constant is added to the numbers $a,$ $b,$ and $c,$ a three-term geometric sequence arises. If $a = 60,$ $b = 100,$ and $c = 140,$ what is the common ratio of the resulting sequence?
+1926 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! :)
