Suppose you were given a four term geometric series such that the four numbers added to
220.
If the difference between the third term and the first term is 44, can you find all four
terms ? Pause the video and see if you can solve this tricky series!
Since the terms of a geometric series have a common ratio, which I’ll call r, we can
replace these four variables with just two.
So our series would look like this.
But one equation with two unknowns isn’t going to get us very far.
We need the other clue to the problem: the difference between the third term and the
first term is 44.
So we can write a times r squared minus a equals 44.
Nice – two equations with two unknowns.
Let’s put the bottom equation on standby and focus on the equation equal to 220.
Let’s first factor out an a.
Then divide both sides by a.
Now let’s factor by grouping by pulling out a factor of r squared from r squared and
r cubed.
So now 1 plus r is a common factor and that allows me to rewrite the left side as this.
Let’s put this equation on standby and do some factoring with the equation equal to
44 now.
Just like before, I’ll first factor out an a.
And then divide both sides by a.
r squared minus 1 is a difference of two squares so we’ll rewrite it as r plus 1 times r
minus 1.
And now I’m going to multiply both sides by 5.
Why did I do that?
Because our previous equation is also equal to 220 over a.
So now I’m going to set these two equations equal to each other and have one equation
in terms of r.
Let’s cancel out the 1 plus r factors from both sides of the equation.
Since this is a geometric series, we can do this without worrying about dividing by zero
since r can’t equal negative 1.
So now we’re left with this.
It looks like we’ve simplified things down to a quadratic equation.
After distributing the 5 we can bring all terms to the left and simplify some more to
get our quadratic in standard form.
This is easily factorable as r minus 3 times r minus 2.
Which means r is equal to 3 and 2.
Let’s start with r=2 and see what our term values are.
Recall that earlier we had the following equation.
Now let’s solve it for a.
And let’s plug 2 in for r.
We can calculate a to be 14 and 2/3.
And if we just multiply 14 and 2/3 by r=2, we’ll get our second term to be 29 1/3.
Again multiplying by 2, we’ll get 58 2/3 for the third term, and finally we’ll get
117 1/3 for the fourth term.
And sure enough, they do add up to 220!
For the second solution we got, r=3.
I’ll leave it to you to confirm that for this solution the first term a is 5.5 and
that the next three terms would be 16.5, 49.5, and 148.5.
All four also add up to 220.
Pretty neat that there are actually two solutions.
This makes sense since the smaller starting value of 5.5 requires a bigger ratio to get
to 220, whereas the larger starting value of 14 and 2/3 required a smaller ratio to
get to 220.
Let me know in the comments below if you got the solutions and how you got them.
And be sure to subscribe for a new video every week featuring puzzles, problems, applications,
and explorations.
Thanks for watching and see you soon!