# Arithmetic Mean & Geometric Mean

Question 1: if the arithmetic mean of two numbers is twice of their geometric mean, their ratio of sum of numbers to the difference of numbers equals?

Question 2: if the quadratic equation:

$(b^2+c^2)x2-2(a+b)cx+(c^2+a^2)=0$

has equal roots then?what is its AP & GP?

Question 3: If the expansion of:

$(1+x)^{50}$

let S be the sum of the coefficient of the odd power of x, then S will be?

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## migrated from mathematica.stackexchange.comOct 2 '12 at 22:13

This question came from our site for users of Mathematica.

This is a site for Mathematica users. You need the Mathematics site – rm -rf Oct 2 '12 at 22:13
What are the definitions of AP and GP? – robjohn Oct 2 '12 at 22:35
What are the AP and GP of a quadratic equation? – Ross Millikan Oct 2 '12 at 23:40
AP is Arithmetic Progression & GP is Geometric Progression. – user1575536 Oct 3 '12 at 0:01

$1.$ We can even do it without the quadratic formula. We have $a+b=4\sqrt{ab}$ and therefore
$$(a+b)^2=16ab.$$ Also, $$(a-b)^2=(a+b)^2-4ab=(a+b)^2-\frac{1}{4}(a+b)^2=\frac{3}{4}(a+b)^2.$$ Thus $$\frac{(a+b)^2}{(a-b)^2}=\frac{4}{3},$$ and therefore $$\frac{a+b}{a-b}=\pm\frac{2}{\sqrt{3}}.$$

$2.$ The roots are equal precisely if the discriminant is $0$, that is, if $$4(a+b)^2c^2-4(b^2+c^2)(c^2+a^2)=0.$$ Divide by $4$, expand everything, do the obvious cancellations. We get $$2abc^2=c^4+a^2b^2,$$ which can be rewritten as $$(c^2-ab)^2=0.$$ We conclude that $ab=c^2$. We cannot have $c=0$ and $b=0$, else we would not have a quadratic equation. We conclude that the sequence $b,c,a$ is a three-term geometric sequence. If $a\ne 0$, then $a,c,b$ is also a three-term geometric sequence.

$3.$ I recommend that you look at the solution by robjohn.

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robjohn & Andre Nicolas --- Thanks for the help. – user1575536 Oct 3 '12 at 0:07

Question 1:

Arithmetic mean: $\frac{a+b}{2}$

Geometric mean: $\sqrt{ab}$

So the condition becomes $\frac{a+b}{2}=2\sqrt{ab}$. Square both sides to get $$\frac{a^2+2ab+b^2}{4}=4ab$$ which, assuming $b\ne0$, results in $$\left(\frac ab\right)^2-14\frac ab+1=0$$ and $$\frac ab=7\pm4\sqrt{3}$$ Then we can compute $$\frac{a+b}{a-b}=\frac{\frac ab+1}{\frac ab-1}=\frac{4\pm2\sqrt{3}}{3\pm2\sqrt{3}}\frac{3\mp2\sqrt{3}}{3\mp2\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}$$

Question 3:

The sum of all the coefficients is $(1+1)^{50}=2^{50}$

The sum of the even coefficients minus the sum of the odd coefficients is $(1-1)^{50}=0^{50}$

The sum of the odd coefficients is $\frac12(2^{50}-0^{50})=2^{49}$.

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if the quadratic equation:

$(b^2+c^2)x^2-2(a+b)cx+(c^2+a^2)=0$

Now we can write it as $\displaystyle (bx-c)^2+(cx-a)^2 = 0$

Means $bx-c = 0\Rightarrow bx = c$ and $cx-a = 0 \Rightarrow cx=a$

So $\displaystyle x = \frac{c}{b} = \frac{a}{c}\Rightarrow c^2 = ab$

So $a,b,c$ are in $\bf{Geometric\; Progression.}$

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