Fizzy AP - GP

Algebra Level 3

Let $a, b \in R^{+}.$

$a, A_{1}, A_{2}, b$ is an arithmetic progression.
$a, G_{1}, G_{2}, b$ is a geometric progression.

Which of the following must be true?

A_{2}G_{1} \geq G_{2}A_{1}

A_{1}G_{2} \geq G_{1}A_{2}

A_{1}A_{2} \geq G_{1}G_{2}

A_{1}A_{2} > G_{1}G_{2}

3 solutions

Sanchayapol Lewgasamsarn
May 15, 2014

$A_{1} = (\frac{b-a}{3}) + a$

$A_{2} = (\frac{2(b-a)}{3}) + a$

$A_{1}A_{2} = ab + \frac{2(b-a)^{2}}{9}$

$G_{1} = a(\frac{b}{a})^{\frac{1}{3}}$

$G_{2} = a(\frac{b}{a})^{\frac{2}{3}}$

$G_{1}G_{2} = ab$

$A_{1}A_{2} - G_{1}G_{2} = \frac{2(b-a)^{2}}{9} \geq 0$

Thus, $A_{1}A_{2} \geq G_{1}G_{2}$

The rest 3 choices may be false for some values of $a, b$

Cant this be done in a little easier way using the A.M-G.M inequality?

Krishna Ar - 7 years ago

Please elaborate.

Sanchayapol Lewgasamsarn - 7 years ago

Sure. Why can't we take A1 and A2 as arithmetic means inserted in this sequence and G1 and G2 as geometric means inserted and say that using the AM-GM inequality A>=G?

Krishna Ar - 7 years ago

@Krishna Ar – But then you'll not get the expressions given in the question.

Satvik Golechha - 7 years ago

what i did was considering a as 1 and common ratio of GP as r and d as common difference of AP and found r in terms of d by equating the 4th terms i.e. b, and found AMs and GMs and eliminated options and got wrong answer

Murlidhar Sharma - 7 years ago

but equal to is possible only when b=a , since a, A1, A2, b are in AP nd a, G1, G2, b are in GP therefore 'a' cannot be equal to 'b'. Hence, A1A2 > G1G2 only !! there is no chance of equality !! therefore, required solution is A1A2 > G1G2 !!

Anish Sony - 7 years ago

Can anyone give me any example where A1A2=G1G2? Cause i still haven't managed to find 1 with both equal. I feel it should be A1A2>G1G2. Please correct me if i am wrong.

Ashu Dablo - 7 years ago

From the properties of parabola, the minimum will be reached when $b-a= 0$ Thus, $a = b$

When $a = b$ , the first and the fourth term of an arithmetic progression will be equal. This can only happen when the common difference is $0$ . An example of this sequence is $a, a, a, a$

When $a = b$ , the first and the fourth term of a geometric progression will be equal. This can only happen when the common ratio is $1$ . An example of this sequence is $b, b, b, b$

When $A_{1}A_{2} = a^{2} = b^{2} = B_{1}B_{2}, Q.E.D.$

Sanchayapol Lewgasamsarn - 7 years ago

but will we consider it as an AP or as GP then??

Ashu Dablo - 7 years ago

equal to is possible only when b=a , since a, A1, A2, b are in AP nd a, G1, G2, b are in GP therefore 'a' cannot be equal to 'b'. Hence, A1A2 > G1G2 only !! there is no chance of equality !! therefore, required solution is A1A2 > G1G2 !!

Anish Sony - 7 years ago

A little different approach. From AP,.. b = a + 3d..<..>

from GP,.. b = ar^3 .. <.....>r^3 = 1 + 3d/a........

.So r=1,..d=0....OR...r > 1.
A1 * A2 = a^2 + 3 * a * d + 2 * d^2......<......>..
G1 * G2 = a^2 * r^3 { from above} = a^2 * ( 1 + 3d/a) = a^2 + 3 * a * d.

If r = 1.. and d = 0, A1 * A2 = G1 * G2...<...>

OR...<...> .. A1 * A2 > G1 * G2.

Niranjan Khanderia - 7 years ago

Α1=2α-β/3 , Α2=2β-α/3 ,G1G2=ab and since agreater than b SO A1A2IS GREATER THAN G1G2

aliki patsalidou - 7 years ago

You're wrong buddy...., (B) is the correct option

Parth Patel - 6 years, 12 months ago

the opt is changing

math man - 6 years, 12 months ago

But A1 A2 = G1 G2 is only true for a=b. But then neither a,A1,A2,b gives an Arithmetic Progression nor does a,G1,G2,b give a Geometric Progression. Did I missunderstand something?

Luis Richter - 1 year, 10 months ago

a common difference of 0 and a common ratio of 1 makes a = b possible.

Kenji Gunawan - 1 year, 9 months ago

Pinak Wadikar
May 21, 2014

I did it using A.M. G.M. Property but my method is a bit different from Arya.

$a+b = A_{1} + A_{2}$

also

$2 \sqrt{A_{1}A_{2}} <= A_{1} + A_{2}$

Similarly,

$ab = G_{1}G_{2}$

And

$2\sqrt{ab} <= a + b$

using these equations we get the answer.

Arya Samanta
May 20, 2014

Let me just put forward how I did using AM-GM Inequality . Also I think its just what Krishna Ar thinks.

We can think (I don't know if its right or wrong) of it like $A_1$ be the arithmetic mean of $a , A_2$ , also $G_1$ be the geometric mean of $a , G_2$ .

Then it stand like $A_1 \geq G_1$ .................................................................................................................( 1 )

Also , I mean Similarily, $A_2 \geq G_2$ .................................................................................................................( 2 )

Multiplying both do we get $\boxed {A_1A_2 \geq G_1G_2}$ ............................................................................................??????( ? )???????

Well that's MY QUESTION. :)

but u cant be so sure about A 2 >= G 2 for all positive values of x

Murlidhar Sharma - 7 years ago

No. Take the rhs in both cases negative and greater in magnitude then the lhs. This is possible for all posiitive real numbers and also if one of the rhs sides is negative and both the lhs are positive .

aneesh kejariwal - 7 years ago

Fizzy AP - GP

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