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?
Please help with this problems in brief. -Thanks.
On
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 $\frac12(2^{50}-0^{50})=2^{49}$.
$1.$ 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.