<div class="text-justify">Suppose we travel back and forth from Lhokseumawe to North Sumatra, The "going" trip is done at a rate of 15 km / h, while the journey "goes home" at a rate of 30 km / h. Then the question arises, "What is the average rate of back and forth travel?"
With great confidence, maybe we immediately replied, (15 + 30) / 2 = 22.5 km / h.

#### Is it true?
It turns out the answer was wrong! Let's examine again. Many of us say that because each rate component has the same "weight" to the average rate. It should be realized that not always the components of the average scorers all have the same weight. Like the question of this average rate, two different rates produce different travel times.

For the same distance, the rate of 30 km / h has a shorter travel time than 15 km / h. The correct answer to this average rate question is to see the value of 15 km / h having a weight 2 times greater than 30 km / h because the travel time is twice as long. Thus, the true average rate is:

<div class="pull-right">
    <center>
        <img src="https://steemitimages.com/DQmNurj5W6N3gMgeyawGz6dBG1d8LGaNgwVpiDSEEXnms51/IMG_20180421_130512.jpg"/>
        <br/>
        <em><a href="http://source.com/link-to-page">My image</a></em>
    </center>
</div>

If we are not so sure of the answer as that, try to review another example that is more "populist". Suppose there is a student who follows ten physics exam in one semester. Nine of the tests he followed were worth 100, while one remaining test was worth 50. Is it fair that the average student score is (100 + 50) / 2 = 75? Certainly not!

The calculation of a fair average value for the student is by first multiplying each value against the weight of each, ie how much the value is obtained. So, the average value of the student is:


<div class="pull-right">
    <center>
        <img src="https://steemitimages.com/DQmadzwgaX2VQwy6Qu1n22fBcZC3GwTBUK9xdj3gKrxUd5G/IMG_20180421_130544.jpg"/>
        <br/>
        <em><a href="http://source.com/link-to-page">My image</a></em>
    </center>
</div>

Now we return to the problem of the average rate. A person who has a high curiosity may ask,
"What if one of those rates is not a multiple of the other?"

Actually, it is easy if you want to be returned to the definition of the average rate, which is the total distance per total time. That is, we the search first time for each rate, use the distance Sabang-Merauke, then divided by the total time.

Maybe the problem is that we are lazy to measure the distance and time of the trip, anyway the data we have is only data rate (or any data value in one group of the same magnitude).


There is a more efficient way to calculate the average rate. Here we will use the concept of harmonic mean, that is, the average of harmonic sequences. The term "harmonic" may be derived from the pattern of numbers: 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, and so on.

If we make a guitar with a relative length of one string with another such string, then will emerge the strains of beautiful melodies harmonious.

avoid it, once we know that we are wanting to calculate the average rate (by means of average harmonics), then we will have a very agreeable formula in calculating the harmonic mean for the same distances.


In the problem we discussed earlier, the distance traveled by each rate is the same, ie the distance between Sabang and Merauke. Below is given directly the formula going, try your own count.

For two rates a and b, the average value formula is:



<div class="pull-left">
    <center>
        <img src="https://steemitimages.com/DQmfTFS1FYpbYanLTU7mmKZ9FBrgAPeKvrmYoCKtm5tBmyC/IMG_20180421_130632.jpg"/>
        <br/>
        <em><a href="http://source.com/link-to-page">My image</a></em>
    </center>
</div>

***

### For three rates a, b, c:



<div class="pull-left">
    <center>
        <img src="https://steemitimages.com/DQmWnybmgcLH1t7RkSM46y5BQ3AWnvJGkmH4HbJhijyC9tu/IMG_20180421_130645.jpg"/>
        <br/>
        <em><a href="http://source.com/link-to-page">For four rates a, b, c, d: My image </a></em>
    </center>
</div>

***
### For four rates a, b, c, d:


<div class="pull-center">
    <center>
        <img src="https://steemitimages.com/DQmYjMfNZDV1HkK3CZqrfe8DGrjg6JhYKj7ks3inEkbpjuC/IMG_20180421_130722%20(1.jpg)"/>
        <br/>
        <em><a href="http://source.com/link-to-page">For four rates a, b, c, d: My image</a></em>
    </center>
</div>


***

So did so. The pattern is very clear, is not it? Apply this average harmonic formula to the initial problem we give, for two rates a = 15 km / h and b = 30 km / h:

<div class="pull-left">
    <center>
        <img src="https://steemitimages.com/DQmSkgfayiB5hJJeg6eryzyVGpxk5VNnMvPpmbxXmeSD5Xn/IMG_20180421_130746.jpg"/>
        <br/>
        <em><a href="http://source.com/link-to-page">My image</a></em>
    </center>
</div>


***

In order to be better, we try another problem that is more difficult.
Usually, a city transport travels back and forth from Lhokseumawe to Cunda then back again from Cunda to Lhokseumawe at an average rate of 300 km/ hour.

The next day, there was a wind that blew at 50 km / h from Lhokseumawe to Cunda. If we assume that the behavior of the driver of transportation remains the same, how is the average rate of travel back and forth on that day after the wind? Is it faster? Slower? Or just the same?

With a passing thought without using the average harmonic formula, we might think the same. The assumption is that the transport received wind assistance on the Lhokseumawe to Cunda route (so the speed is 350 km / h), while on the Cunda Ke Lhokseumawe line gets the wind "obstacle" (so the speed is 250 km / h). With the wrong mindset as at the beginning of this paper, the average rate is obtained (350 + 250) / 2 = 300 km /hour. The result is at a glance the same. But try to apply the average harmonic formula,


<div class="pull-center">
    <center>
        <img src="https://steemitimages.com/DQmUKbCZKAX2LQfMa93eHs9WqYs9J3RgH3ZHBT8qWcZrGsw/IMG_20180421_130804.jpg"/>
        <br/>
        <em><a href="http://source.com/link-to-page">My image</a></em>
    </center>
</div>

***

###### It turns out to be slower! And this is the right result. Also try the average harmonic formula for other data that has more value, not just two values. Have fun.</div>


***
## Thank for taking the time to read my post




https://steemitimages.com/DQmSFYqSzFWNsB3FWh5ZtLVyeKVsbUkJynjHsUAVsjQaRsC/LOGO%20STEEMIT%20SWARD%20LEBIH%20OK.jpg

***