Question 51321: The value of xyz is 15/2 or 18/5 accordingly if the series a,x,y,z,b is arithmetic progression or harmonic progression.Find a,b.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! The value of xyz is 15/2 or 18/5 accordingly if the series a,x,y,z,b is arithmetic progression or harmonic progression.Find a,b.
THERE IS SOME AMBIGUITY IN WRITE UP..PLEASE CHECK UP THE PROBLEM AGAIN...
SAME X,Y,Z ARE MENTIONED IN BOTH CASES A.P...OR...H.P?THEN HOW CAN THEIR PRODUCT BE DIFFERENT 15/2 AND 18/5... ARE THEY DIFFERENT SAY X,Y,Z IN ONE CASE AND X',Y',Z' IN ANOTHER CASE ??
OK..LET ME CORRECT YOUR WRITE UP TAKING THEY ARE 3 DIFFERENT NUMBERS X,Y,Z AND X',Y',Z'..
CASE 1......A,X,Y,Z,B...ARE IN A.P.....N=5....LET D BE THE C.D
B= A+4D......
(B-A)/4=D
SO X,Y,Z ARE [A+(B-A)/4],[A+2(B-A)/4],[A+3(B-A)/4]
THEIE PRODUCT = 15/2 = (3A+B)(A+B)(A+3B)/32
(3A+B)(A+B)(A+3B)=32*15/2=240..................1
CASE 2......A,X',Y',Z',B ARE IN H.P.
SO....1/A,1/X',1/Y',1/Z',1/B ARE IN A.P....N=5....LET D'=C.D
1/B = 1/A + 4D'
D'=(A-B)/4AB
SO 1/X',1/Y',1/Z' ARE....AS ABOVE ...WE GET ...(A+3B)/4AB , (A+B)/2AB , (3A+B)/4AB
SO....
X'Y'Z'=32A^3B^3/(A+3B)(A+B)(3A+B)=18/5.................2
EQN.1*EQN.2....
32A^3B^3 = 240*18/5
A^3B^3=240*18/(5*32)=27
AB=3.................................3
FROM EQN.1 AND EQN.3 ,WE SEE BY INSPECTION THAT A=1 AND B=3..SINCE AB=3 AS PER EQN.3 AND (1+3*3)(1+3)(1*3+1)=10*4*6=240 AS PER EQN.1..
SO THE 2 NUMBERS ARE
A=1...AND...B=3.OR...........A=3 AND B=1...
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