Ask Uncle Colin: A Weird Arithmetic Progression

Dear Uncle Colin,

I'm told that the three terms $a_1 = \log(2)$, $a_2 = \log(2\sin(x)-1)$ and $a_3 = \log(1-y)$ are in arithmetic progression and I need to find the range of possible values for $y$. I don't really know where to start!

- Logarithmic Arithmetic Progression Lacks A Clear Explanation

Hi, LAPLACE, and thanks for your message!

If three terms are in arithmetic progression, it means that the differences between consecutive terms is always the same. That is to say:

$\log(2\sin(x)-1) - \log(2) = \log(1-y) - \log(2\sin(x)-1)$ (*)

There are further restrictions on what can go into $\log$ as an argument: $1-y$ and $2\sin(x)-1$ must both be positive. Put another way, $y < 1$ and $\sin(x) > \frac{1}{2}$.

Rearranging (*) and getting rid of logs gives us $(2\sin(x)-1)^2 = 2(1-y)$. However, we know that $(2\sin(x)-1)$ is greater than zero and - by definition - no greater than 1. That means $0 < (2\sin(x)-1)^2 \le 1$ - and the middle of that inequality could just as well be $2(1-y)$, because they're equal.

$0 < 2(1-y) \le 1$

That rearranges to $-1 < -y \le -\frac{1}{2}$, which means that $\frac{1}{2} \le y \lt 1$.

Double-checking, that satisfies the original inequality we had for $y$, so it's our final answer.

Hope that helps!



Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


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