Question

Match the following

Column I	Column II
(A) If $a,b,c$ be positive numbers then $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ must be greater than or equal to	$9$
(B) If $h$ be the H.M and $g$ be the G.M of two positive numbers $a$ and $b$ such that $h:g=4:5$ then $\frac{a}{b}$ can be equal to	$4$
(C) If $S={\sum }_{r=0}^{\infty }\frac{1}{2^r}$ and $S_{n+1}={\sum }_{r=0}^n\frac{1}{2^r}$ and $S-S_{n+1}<{10}^{-3}$ then $n$ is greater than or equal to	$10$
(D) If $(1+x)(1+x^2)(1+x^4)(1+x^8)...(1+x^{128})={\sum }_{r=0}^n{x}^r$ then $n$ is equal to	$255$

Answer 1

$A)$ As we can see the least value is a constant,

ie, the value of $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ doesn't depend on $a,b$ or $c$

So, $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ is least when $a=b=c=k$

$\Rightarrow \left(3k\right)\left(\frac{3}{k}\right)=9$

$\left(A\right)\to 9$

$B)$ $h=\frac{2ab}{a+b}$ and $g=\sqrt{ab}$

$\Rightarrow \frac{h}{g}=\frac{2\sqrt{ab}}{a+b}=\frac{4}{5}$

$\Rightarrow \frac{(a+b{)}^2}{ab}=\frac{25}{4}\Rightarrow \frac{(a+b{)}^2-4ab}{ab}=\frac{25}{4}-4=\frac{9}{4}$

$\Rightarrow \frac{(a-b{)}^2}{ab}=\frac{9}{4}$

So, $\frac{(a+b{)}^2}{(a-b{)}^2}=\frac{25}{9}\Rightarrow \frac{a+b}{a-b}=\frac{5}{3}$

On doing componento-dividento,

$\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{a}{b}=\frac{5+3}{5-3}=\frac{8}{2}=4$

$\left(B\right)\to 4$

$C)$ $S={\sum }_{r=0}^{\infty }\frac{1}{2^r}=\frac{1}{1-\frac{1}{2}}=2$ (infinite sum of G.P with $r=\frac{1}{2}$ and $a=1$)

$S_{n+1}=\frac{1-(\frac{1}{2}{)}^{n+1}}{1-\frac{1}{2}}=\frac{2^{n+1}-1}{2^n}$

$S-S_{n+1}=2-\frac{2^{n+1}-1}{2^n}=\frac{1}{2^n}=2^{-n}<{10}^{-3}\Rightarrow 1000<2^n$

$\Rightarrow n\ge 10$

$\left(C\right)\to 10$

$D)$ $\frac{(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{128})}{1-x}=\frac{1-x^{256}}{1-x}={\sum }_{r=0}^{255}{x}^r$

So, $n=255$

$\left(D\right)\to 255$