Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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

Question 3: If the expansion of:


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

Please help with this problems in brief. -Thanks.

share|cite|improve this question

migrated from Oct 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
up vote 1 down vote accepted

$1.$ 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.

share|cite|improve this answer
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 $\frac12(2^{50}-0^{50})=2^{49}$.

share|cite|improve this answer

if the quadratic equation:


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 $\bf{Geometric\; Progression.}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.