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Match the following.
LIST-I
LIST-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
P)
4
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 a/b can be equal to
Q)
9
C)
If $S = \sum_{r = 0}^{\infty} \frac{1}{2^{r}}$ and $S_{n} = \sum_{r = 0}^{n} \frac{1}{2^{r}}$ and $S - S_{n} > 10^{- 3}$ then n is less than or equal to
R)
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
S)
255

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$A \to P, Q; B - P; C - Q; D - S$

$A \to P, Q; B - P; C - Q; D - P$

$A \to S; B - P; C - Q; D - P, Q$

$A \to P, Q; B - S; C - Q; D - P$

answer is B.

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Detailed Solution

A: $a + b + c \geq 3 \sqrt[3]{a b c}$
and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 3 \sqrt[3]{\frac{1}{a b c}}$
$\Rightarrow (a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$
B: $\frac{2 a b}{a + b} \times \frac{1}{\sqrt{a b}} = \frac{4}{5}$
$\Rightarrow 5 \sqrt{a b} = 2 (a + b)$
$\Rightarrow 25 a b = 4 a^{2} + 4 b^{2} + 8 a b$
$\Rightarrow 4 {(\frac{a}{b})}^{2} - 17 (\frac{a}{b}) + 4 = 0$
$\Rightarrow (\frac{a}{b} - 4) (4 (\frac{a}{b}) - 1) = 0$
$\Rightarrow \frac{a}{b} = 4$ or $\frac{1}{4}$
C: $S = \frac{1}{1 - \frac{1}{2}} = 2$
Now, $2 - 2 (1 - \frac{1}{2^{n + 1}}) \geq 10^{- 3}$
$\Rightarrow \frac{1}{2^{n}} > 10^{- 3}$
$\Rightarrow 1000 > 2^{n}$
$\Rightarrow n < 10$
D: Multiply and divide by $(1 - x)$
$(1 + x) (1 + x^{2}) .............. (1 + x^{128}) = \frac{1 - x^{256}}{1 - x}$
$= 1 + x + x^{2} + 1........... + x^{255}$

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