IIT JEE , AIEEE Forum - Mathematics , Algebra

neeraj_agarwal_1990

If m is the root of the line equation (1-ab)x² - (a²+b²)x - (1+ab)=0 and m harmonic means are inserted between a and b ,then the difference between the last and the first of the means equals?
[ans:ab(b-a)]

pink_ele

are you confirmed about the ques please see the equation

amaron

(1-ab)x² - (a²+b²)x - (1+ab)=0

let the common difference of the reciprocals of the HP,which will be in AP be d.

then 1/b = 1/a + (m+1)d

implies d =(a-b)/(ab(m+1)).   ~~~~~~~~~~(1)

as m is a root of (1-ab)x² - (a²+b²)x - (1+ab)=0 ,

(1-ab)m² - (a²+b²)m - (1+ab)=0

implies    m² - 1 = m²ab + (a²+b²)m +ab

implies    m² - 1 = (bm+a)(am+b),implies

(m² - 1)/(bm+a)(am+b) =1     ~~~~~~~~~~(2)

now first harmonic mean = 1/(1/a +d ), substituting value of d from (1),

first harmonic mean = 1/(1/a + (a-b)/(ab(m+1))) = (ab(m+1))/(bm + a),
similarly,
last harmonic mean= 1/(1/b - d) = (ab(m+1))/(am + b)
implies,
difference of first and last harmonic means =

(ab(m+1))/(bm + a) - (ab(m+1))/(am + b) =

ab(m+1)(b-a)(m-1)/(bm+a)(am+b) = ab(b-a)(m² - 1)/(bm+a)(am+b)

but from (2), (m² - 1)/(bm+a)(am+b) =1

therefore,

difference of first and last harmonic means = ab(b-a)*1 =ab(b-a).

That solves the problem!

amar.gupta

good work amaron

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