IIT JEE , AIEEE Forum - Mathematics , Algebra

himanshu2006

IF THE +VE REAL NO.S a,b,c are in A.P. with abc=4 , then minimum value of

b is

a) 1 b) 3 c) 2 d) 1/2

shakirshafi12

a,b,c are in A.P
therefore d is the common difference;
abc=4;
A.M>=G.M
(a+c)/2>=sqroot(ac)
in an a.p (a+c)/2=b;
b>=sqroot(4/b)
for minima b=(4)^1/3
i.e 1.5874
experts plzzzz reply i am not sure of this answer

ace

is taking of gm for no.not in gp allowed

hit_ur_heart

i think shafi is correct ,

it is like this , since a , b , c are positive real numbers , therefore they will satisfy the fact that A.M . >= G.M

=> a+b+c/3 >= (abc)^(1/3)
substituting abc = 4 && 2b = a+c
we get ,
b >= (4)^(1/3)

ace

but shafi if we go by ur answer wont we get 2 answers

shakirshafi12

how??????
i have used
graphical,A.M>G.M,maxima minima approach all lead to same result
but b min will occur only when d=0

ace

real smart man

mohit1

shakirshafi can you tell me where do you study?

shakirshafi12

Amity International school
M-block Saket
New Delhi
why did you ask me?

mohit1

i mean where do you take coaching from?

iit2007

by A.M>G.M

minimum value of b is 2

rashmi_jain

A.M. >=G.M.

(a+b+c)/3>= (abc)^1/3

a,b,c are in AP therefore,a+c=2b

therefore,(2b+b)/3>= 4^1/3

b>=4^1/3

so minimum value of b is 4^1/3

but this is not there in the options.so,the number greater than this should be the answer i.e.2

Asmita

Since a, b, c are in A. P. Therefore B is the A.M. of a,c.

By A.M.-G. M. inequality,

(a+c)/2 >=

ac

b>=

(4/b) (abc=4 )

squaring,b²>=4/b

as b is a +ve real no. .^..b³>=4

for b to be minimum, b³ = 4

b=³

4 = 2^2/3

Ask & Discuss Questions with Community & Experts