Here is a textbook situation in Time, Speed and Distance: A man goes from point A to point B with velocity *v*_{1} and returns with velocity *v*_{2}. What is his average velocity?

Using the formula , the average velocity can be found to be . So far so good.

Ever wondered why we get this result? Why do we get the velocity as the harmonic mean of the two velocities? The answer lies in basics of arithmetic and harmonic progressions.

Let *t*_{1} and *t*_{2} be the time taken while going from A to B and coming back. The situation is shown below:

Now what will a hypothetical average velocity in this situation mean? It will mean that a person takes the same time, *t*_{average}, while going from A to B and coming back. The situation is summarized below:

The total time taken will be same as *t*_{1} + *t*_{2}

--> 2 × *t*_{average} =* t*_{1} + *t*_{2} or *t*_{average} = (*t*_{1} + *t*_{2})/2

In another words, ** t_{1}, t_{average}, and t_{2} will be in arithmetic progression**.

So how is this related to velocity?

Remember that when distance is constant velocity is inversely proportional to time?

I.e. V is proportional to 1/T or V = k/T.

If T_{1}, T_{2}, and T_{3} were in **arithmetic progression**, then 1/T_{1}, 1/T_{2}, 1/T_{3} are in **harmonic progression** => k/T_{1}, k/T_{2}, k/T_{3} are in **harmonic progression =>** V_{1}, V_{2} and V_{3 }are in **harmonic progression!**

It can also be proved that if V_{1}, V_{2} and V_{3 }are in **arithmetic progression**, then T_{1}, T_{2}, and T_{3} are in **harmonic progression.**

So here’s the rule:-

Now let’s apply these rules in practical problems:-

Re: Quant Exotica- Use of Arithmetic and Harmonic Progressions in Time, Speed, and Distance | |

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