IIT JEE , AIEEE Forum - Mathematics , Algebra

umang

if (a+b)/(1-ab) , b , (b+c)/(1-bc) are in AP , then a , 1/b , c are in ???

!. AP

2. GP

3. HP

4. None

snehavenus

hi umang bhaiya.........the answer is H.P

since a+b/1-ab, b, b+c/1-bc are in A.P

(b+c) / (1-bc) - b = b- (a+b)/(1-ab)

this implies (b+c-b+b²c) / (1-bc) = (b-ab²-a-b) / (1-ab)

c(1+b²)/(1-bc) = -(a(1+b²))/ (1-ab)

c-abc= -a+abc this implies 2abc=c+a

2b= (c+a)/ ac = 1/a+1/c this again implies 1/a, b , 1/c are in

A.P and hence a, 1/b , c are in H.P

pls tell me if i am wrong.........

vineetvsb

ialso agree with him
Since (a+b)/I-ab,b,(b+c)/1-bc are in a.p.
Therefore2b=[(a+b)/I-ab]+[ ,(b+c)/1-bc]
=2b=(a-abc+b-b^2c+b+c-abc-ab^2)/(1-ab^2c-bc-ab)
2b-2ab^3c-2b^2c=a+2b+c-2abc-b^2c-a^2
=2ab^3c-b^2c-ab^2=a+c-2abc
=-(b^2c+ab^2)=a+c-2abc+2ab^3c
= -b^2(a+c)=[a+c-2abc(1+b^2)]
=-b^2(a+c)-(a+c)= -2abc(1+b^2)
= b^2(a+c)-(a+c)= 2abc(1+b^2)
=(a+c)(b^2+1)= 2abc(1+b^2)
=(a+c)=2abc
=(1/b)=2ac/(a+c)
therefore a,1/b,c are in h.p.

plz.........rate me

umang

Thanks sneha !!!

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