If positive numbers a, b, c are in arithmetic progression such that ab

Question

If positive numbers a, b, c are in arithmetic progression such that abc = 4, then the minimum value of b is

A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)

If positive real numbers a,b,c are in a.p such that abc=4 Then the minimum value of b is

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Nidzo · Accepted Answer

We are told,  abc = 4 . Looking at the answer choices all are exponents of 2, so let  abc = 2^2

As we are dealing with an arithmetic sequence we know that each term in the sequence will have the same base and the difference between them each will be a constant. Let

Let  a = 2^n ,  b = 2^{n+x}  &  c=2^{n+2x}

Therefore  (2^n)*(2^{n+x})*2^{n+2x} = 2^2

2^{n+n+x+n+2x} = 2^2

Drop the bases

3n + 3x = 2

What we know is that  b = 2^{n+x}  so to get  n+x  we divide the above equation through by 3 to get:

n+x = \frac{2}{3}

So  b = 2^{\frac{2}{3}}

Answer B

chetan2u · Answer

kapil1995 , I believe the solution above is not correct as  2^n,2^{n+x},2^{n+2x}  is  2^n,2^x(2^n),2^x(2^x2^n) , that is successive term is getting multiplied by  2^x , making the sequence Geometric progression.

Logical :
a*b*c=4 and looking at the options 
Since the terms are equidistant, the way to minimise b and c as also maximise a is by taking all three same => a=b=c
Thus,  a*b*c=b^3=4=2^2………b=2^{\frac{2}{3}}

Algebraic method

Now, you could also verify that we have to look for  b^3=2^2 
Let the product be  (b-x)*b*(b+x)=b(b^2-x^2)

Now  b^2-x^2<b^2  unless x is 0.
If x can be 0 too, then  b^2-x^2\leq b^2

Multiply both sides by b without changing the sign   as b is positive  
 b(b^2-x^2)\leq b(b^2) , but  b(b^2-x^2)=4

Thus,  4\leq b^3……2^2\leq b^3 
Taking cube root,
 2^{\frac{2}{3}}\leq b 
Thus, b is at least  2^{\frac{2}{3}}

Ivy17 · Answer

Hi  Nidzo , can you please help me understand how you inferred that in an AP, each term will have the same base? An AP could be 3, 5, and 7, but seemingly their bases are all diff.

Nidzo · Answer

Ivy17

Looking at the formula for an arithmetic sequence  a_n = a_1 + (n-1)d  ,  a_1  would be what I referred to as the base.

Using your example of 3, 5, 7. The difference between each proceeding value is 2. So we can rewrite it as 
3:   3+(1-1)*2  
5:   3+(2-1)*2  
7:   3+(3-1)*2

Ivy17 · Answer

Ivy17

Looking at the formula for an arithmetic sequence \(a_n = a_1 + (n-1)d \), \(a_1\) would be what I referred to as the base.

Using your example of 3, 5, 7. The difference between each proceeding value is 2. So we can rewrite it as
3: \( 3+(1-1)*2\)
5: \( 3+(2-1)*2\)
7: \( 3+(3-1)*2\)

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ThatDudeKnows · Answer

Since a, b, and c are all positive, the arithmetic mean is greater than the geometric mean.

Arithmetic mean is b.
Geometric mean is cubed-root of 4.

b >= 4^{1/3}

b >= 2^{2/3}

Answer choice B.

Kinshook · Answer

If positive numbers a, b, c are in arithmetic progression such that abc = 4, then the minimum value of b is

2b = (a+c)
abc = 4

b = 4/ac = (a+c)/2

For minimum value of b, ac is to maximised and a+c is to be minimised.
It is possible when a = c

b = a = c
b = 4^(1/3) = 2^(2/3)

IMO B

djangobackend · Answer

just do x-d * x * x+d now b = 4/ac now for b to be min. ac = max so product max when both values equal, (I though AC max will be 4 so we can do a=2, c=2 now if a=2, c=2 then b=1 - but its incorrect since we do not know if values are integers or not) so we do A=C i.e., x-d = x+d so d=0 now basically X^3 = 4 which gives values of X as option B

btsaami · Answer

Since a, b and c are in AP. Let a=b-d, b=b, c=b+d, where d is the common difference.
We know that, AM>=GM
a+b+c/3 >= cube root (abc)
b-d+b+b+d /3 >= cube root (4) or 4^(1/3)
3b/3>=4^(1/3)

Therefore b= 2^(2/3)

bumpbot · Answer

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If positive numbers a, b, c are in arithmetic progression such that ab

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