Harmonic Means |Algebra | Grade 10 Math - YouTube

Today's lesson is Harmonic Means

Harmonic means are terms between  two terms of a harmonic sequence.  

Example. 1 over 5, 1 over 10, 1 over 15, 1 over  20, 1 over 25, and so on is a harmonic sequence.  

The harmonic means between 1 over 5, and 1 over  25 are 1 over 10, 1 over 15, and 1 over 20.

Another example. 1 over 2, 1 over 9, 1 over 16,  1 over 23, 1 over 30, is a harmonic sequence.  

The harmonic means between 1 over 9, and  1 over 30 are 1 over 16 and 1 over 23.

Example number one. Insert the harmonic  mean between one over 15 and one over 37.  

Harmonic mean is equal to 2 over a plus b. This  is the simple formula I would like to share.  

I arrived at this formula by manipulating  algebraically the formula for finding the nth term  

of an arithmetic sequence .I solved for d, the  common difference and then added the result  

to a, which represents the first term, then took  the reciprocal of the result. In this formula,  

a represents the reciprocal of the  first term of the two given terms  

and b represents the reciprocal of the second  term of the two given terms. So a is equal to  

15 the reciprocal of 1 over 15 and b is 37 the  reciprocal of 1 over 37. So HM, HM means harmonic mean  

is equal to 2 over 15 plus 37. This is equal  to 2 over 52. 15 plus 37 is 52. Then simplify,

1 over 26 .So the harmonic mean is 1 over 26.

Example number 2. Insert the harmonic  mean between 1 over 15 and 1 over 37.  

This is the same as the problem in example number  one. Let us solve it by using the usual method.  

Find the reciprocal of 1 over 15  and the reciprocal of 1 over 37.

So we have now 15, the missing term, and then 37.

Write the formula for finding the nth term  of an arithmetic sequence. The formula is  

a sub n is equal to a sub 1 plus the quantity n  minus 1 times d. a sum n refers to the last term,  

a sub 1 the first term, n the number  of terms, d the common difference.

a sub n is equal to 37, a sub 1 is equal to 15,  

n is equal to 3, 1 2 3. We don't know yet the  value of d. Substitute the values in the formula.  

So we have 37 is equal to 15 plus the  quantity 3 minus 1 times d. This is now  

equal to 37 is equal to 15 plus 2 d. 3 minus  1 is 2. Transpose 15 to the left side, change  

its sign from positive to negative. 37 minus  15 is 2 d. 37 minus 15 is equal to 22 and then  

equal 2 d. Divide both sides by 2. So d is equal  to 11. This is the common difference. Add 11 to  

15 to get the second term. So 15 plus 11 is equal  to 26. So 26 is the second term of this sequence.

The reciprocal of 26 is 1 over 26. 1 over 26 is  the harmonic mean between 1 over 15 and 1 over 37.

Example number three. Insert three harmonic  means between 1 over 12 and 1 over 36.

Let us use the formula Harmonic Mean is  equal to 2 over a plus b. There are three  

missing harmonic means here between 1 over  12 and 1 over 36, one two three. Let us first  

find the the second harmonic mean. Our a  is equal to 12 the reciprocal of 1 over 12  

and our b is 36 the reciprocal of 1 over  36. This is now equal to 2 over 12 plus 36.  

12 plus 36 is 48. So this is now equal to 2 over  48. Simplify 2 over 48. So we have 1 over 24.  

So the missing harmonic mean here is 1 over 24.  Next, we are going to find the missing harmonic  

mean here, so Harmonic Mean is 2 over a plus  b. Our a is 12, the reciprocal of 1 over 12  

and our b is 24, the reciprocal of 1 over  24. So this is now equal to 2 over 12  

plus 24. This is now equal to 2 over 36. Simplify  2 over 36. This is now equal to 1 over 18. So the  

missing harmonic mean here is 1 over 18. Next,  we will find the missing harmonic mean here.  

So Harmonic Mean is 2 over a plus b. Our  a now is 24, the reciprocal of 1 over 24  

and our b is 36, the reciprocal of 1 over 36.  So this is now equal to 2 over 24 plus 36.  

This is now equal to 2 over 60. Simply by 2 over  60. We have 1 over 30. So here 1 over 30. The  

three harmonic means between 1 over 12 and 1  over 36 are 1 over 18, 1 over 24 and 1 over 30.

Example number four. Insert three harmonic  means between 1 over 12 and 1 over 36.  

This is the same as the problem in example number  three. Let's solve it by using the usual method.  

Find the reciprocal of 1 over  12 and the reciprocal of 1 over  

36. So we have now 12, a missing term, another  missing term, another missing term, and then 36.  

Next write the formula for finding the nth  term of an arithmetic sequence. The formula is  

a sub n is equal to a sub 1 plus the quantity  n minus 1 times d, where a sub n represents the  

last term, a sub 1 represents the first term, n  the number of terms, and d the common difference.  

a sub n is 36, a sub 1 is 12, n is 5, 1 2  3 4 5. We don't know yet the value of d.  

Substitute these values in the formula. So we  have 36 is equal to 12 plus 5 minus 1 times d.

We have 36 is 12 plus 4d. 5 minus 1 is equal  to 4. Transpose 12 to the left side. Change its  

sign from positive to negative. So we have now 36  minus 12 is equal to 4d. 36 minus 12 is equal to  

24 and this is equal to 4d. Divide both  sides of the equation by 4. So d is equal to  

6. This is the common difference. To find  the second term here, you add 12 to 6. So  

12 plus 6 is equal to 18. To find the next term,  add 6 to 18. So 18 plus 6 is 24 and then 24 plus 6

is equal to 30.  

Next, find the reciprocal of 18, the  reciprocal of 24 and then the reciprocal of 30.

The reciprocal of 18 is 1 over 18. The reciprocal  of 24 is 1 over 24 and the reciprocal of 30 is  

1 over 30. These are the three harmonic  means between 1 over 12 and 1 over 36.

Example number five. Insert two harmonic  means between one over seven and one over  

19. Find the reciprocal of 1 over  7 and the reciprocal of 1 over 19.  

The reciprocal of 1 over 7 is 7 and the reciprocal  1 over 19 is 19. So we have 7, a missing term,  

another missing term, and then 19. The formula  Harmonic Mean is equal to 2 over a plus b can  

be applied only if the missing harmonic mean  is exactly at the middle or exactly between  

two terms of a harmonic sequence. In the problem,  there is no missing harmonic mean that is exactly  

between 1 over 7 and 1 over 19, so we cannot  apply that formula. The formula we are going  

to apply is the formula for finding the nth  term of an arithmetic sequence. The formula is  

a sub n is equal to a sub 1 plus the quantity  n minus 1 times d where a sub n represents  

the last term, a sub 1 the first term, n the  number of terms, and d the common difference.

a sub n is 19,

a sub 1 is 7, n is 4, 1 2 3 4. We don't know  yet the value of d. Substitute the values in  

the formula. So we have 19 is equal to  7 plus the quantity 4 minus 1 times d.

We have now 19 is equal to 7 plus 3 d. 4 minus  1 is three. Transpose seven to the left side.  

Change its sign from positive to negative. We  have now 19 minus seven is equal to three d.  

19 minus 7 is 12, which is equal to 3 d.  Divide both sides by 3. So d is equal to 4.  

Four is the common difference. To find the next  term after seven, add four. So seven plus four is  

eleven. To find the next term  after eleven add four. So  

eleven plus four is equal to fifteen. Next find  the reciprocal of eleven and the reciprocal of 15.  

The reciprocal of 11 is 1 over 11 and  the reciprocal of 15 is 1 over 15.

1 over 11 and 1 over 15 are the two harmonic  means between 1 over 7 and one over nineteen.