..

1. HARMONIC
PROGRESSION
 PRESENTATION OF GROUP 6
 NAME OF PATICEPENTS
 ZOHA AYUB
 SUGRAH RAHIM
 SAWIRA
 FARIHA KARIM AND JIYA MURAD

2. What is harmonic
progression
A HARMONIC PROGRESSION IS A
SEQUENCE OF QUANTITIES WHOSE
RECIPROCAL FORM AN ARITHMETIC
PROGRESSION

3. POINTS TO BE NOTED
THE SERIES FORMED BY
RECIPROCAL OF THE TERMS OF A
GEOMETRIC SERIES
THERE IS NO GENERAL METHOD OF
FINDING THE SUM OF THE
HARMONIC PROGRESSION

4. Examples
THE SEQUENCE
S1,S2……………..SN
IT IS HARMONIC PROGRESSION IF
1/S1,1/S2…………………1/SN
FORM AN ARITHMETIC PROGRESSION

5. MATHOD FOR RE CHEKING A
HARMOINIC PROGRESSION
A HARMONIC PROGRESSION IS SET OF VALUES
THAT ARE ONCE RECIPROCATED RESULT TO AN
ARITHMETIC PROGRESSION TO CHECK THE
RECIPROCATED VALUES MUST POSSES A
RATIONAL COMMON ONCE THIS HAS BEEN
IDENTIFIED WE MAY SAY THAT THE SEQUENCE IS A
HARMONIC PROGRESSION

6. HARMONIC MEAN
HARMONIC MEAN ARE THE TERMS
FOUND IN BETWEEN TWO TERMS OF
HARMONIC PROGRESSING

7. PROBLEMS
DETERMINE WHICH OF THE FOLLOWING ARE HARMONIC
PROGRESSION
1- 1 ,1/2,1/3,1/4
STEP 1= RECIPROCAL ALL THE GIVEN TERMS
THE RECIPROCALS ARE
= 1,2,3,4
STEP 2 = IDENTIFY WEATHER THE RECIPROCATE SEQUENCE IS A
ARITHMETIC PROGRESSION BY CHECKING COMMON DIFFERENCE
BETWEEN THE TERMS
THE ANSWER IS HARMONIC PROGRESSION

8. 2- 1,1/4,1/5,1/7
BY REPEARTING THE STEPS THAT WE HAD DONE
BEFORE WE GET THE RESULT THAT THIS IS NOT
HARMONIC PROGRESSION
3- DETERMINE THE NEXT THREE TREMS OF EACH OF
FOLLOWING HARMONIC PROGRESSIN
1- 24,12,8,6
STEP 1- RECIPROCAL OF FOLLOWING
1/24,1/12,1/8,1/6
STEP 2- FINDING THE COMMAN DIFFERENCE

9. WE CAN FIND COMMON DIFFERENCE THROUGH
SUBTRACTING THE SECOND TREM TO THE FIRST
TREAM ,THIRD TO SECOND AND FOURTH TO THIRD
AND SO ON FOURTH
FOR THE SAME QUESTION WE CAN FIND COMMNON
DIFFERNCE
1/24,1/12,1/8,1/6
=1/12-1/24
=2/12-1/24
=2-1/24
=1/24 ANSWER

10. AS WE HAVE TO FIND THE HARMONIC PROGRESSING FOR
NEXT THREE TERMS SO
1/24 FIRST TERM
1/12 SECOND TERM
1/8 THIRD TERM
1/6 FOURTH TERM
WE HAVE TO FIND FIFTH ,SIXTH AND SEVENTH
SO TO FIND FIFTH TERM
=1/6 WHICH IS FOURTH TERM ADD FROM FIRST TERM
WHICH IS 1/24
=1/6+1/24
=4/24+1/24
FIFTH TERM WILL BE =5/24 RECIPROCAL =24/5

11. FOR SIXTH TERM
5/24 WHICH IS FIFTH TERM ADD FTOM FIRST TERM WHICH
IS 1/24
SO, = 5/24+1/24
=6/24
=1/4 RECIPROCAL =4
FOR SEVENTH TERM
¼ WHICH IS SIXTH TERM ADD FROM FIRST TERM 1/24
SO =1/4+1/24
=6/24+1/24
=7/24 RECIPROCAL =24/7

12. HOW TO FIMD HARMONIC MEAN
BETWEEN THE TREMS
1- 12 AND 8
STEP 1- TAKE RECIPROCAL OF THE TREMS
SO THEY BECOME =1/12 AND 1/8
STEP 2- ARRANGE THE GIVEN TREMS AS FOLLOW
1/12 FIRST TREM
HARMONIC MEAN SECOND TREM
1/8 THIRD TREM

13. FOR FURTHER SOLUTION WE HAVE FORMULH THAT
TN=T1+(N+1)D
WE MAY NOW SUBSTITUTE THE VALUE OF PROBLEMS
TO THE FORMULA TO FIND THE COMMAS DIFFERENCE ( D
) AND THE HARMONIC MEAN IS AS FOLLOW
T3 =T1 +(3-1)D
=1/8 =1/12 +2D
=(3-1)/24=2D
=(3-2)=48D
= 1=48D
= D=1/48

14. AFTER GETTING THE COMMON DIFFERENCE ,ADD IT
TO FIRST TERM TO GET THE HARMONIC MEAN
BETWEEN THE TWO TERMS
T2=T1+D
T2 =1/12+1/48
T2=(4-1)/48
T2 =5/48
RECIPROCAL
T2 =48/5

15. 1- FIND HARMONIC MEAN BETWEEN 36 AND 36/5
SO FIRST RECIPROCAL OF THE TERMS
SO THEY BECOME 1/36 AND 5/36
ARRANGING THE GIVEN TERMS
1/36 FIRST TERM
HARMONIC MEANS SECOND, THIRD AND
FOURTH
5/36 FIFTH TERM
FOR THIS PROBLEM WE HAVE FORMULA
TN =T1+(N-1)D

16. BY PUTTING VALUES
T5 =T1+(5-1)D
= 5/36=1/36 +4D
= 5/36-1/36 =4D
= (5-1)=144D
= 4=144D
= D=4/144
= D = 1/36

17. AFTER FINDING THE COMMON DIFFERENCE KNOW
FIND THE HARMONIC MEAN
T2=T1+D
T2 =1/36 +1/36
T2 =2/36
T2 =1/18
RECIPROCAL
T2 =18

18. T3=T2+D T4=T3+D
T3= 2/36+1/36 T4=3/36+1/36
T3=3/36 T4=4/36
T3=1/12 T4=1/9
RECIPROCAL RECIPROCAL
T3=12 T4=9

19. THERE FORE THE THREE MEAN BETWEEN 36 AND 1/36
ARE 18 ,12AND 9
THANK YOU