If1n-1-131-132-133-1n- then n-

130 gt;133;130 gt;131;130 gt;131; ⇒130+130+130 gt;131+132+133 ⇒330 gt;131+132+133 ⇒110 gt;131+132+133 ⇒ nbsp;1n = 110⇒ n=10 and OR 131 g ...

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If1n+1<131+13...

Question

$I f \frac{1}{n + 1} < \frac{1}{31} + \frac{1}{32} + \frac{1}{33} < \frac{1}{n}; t h e n n ?$

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Solution

The correct option is B 10
$\frac{1}{30} > \frac{1}{33}; \frac{1}{30} > \frac{1}{31}; \frac{1}{30} > \frac{1}{31};$

$\Rightarrow \frac{1}{30} + \frac{1}{30} + \frac{1}{30} > \frac{1}{31} + \frac{1}{32} + \frac{1}{33}$

$\Rightarrow \frac{3}{30} > \frac{1}{31} + \frac{1}{32} + \frac{1}{33}$

$\Rightarrow \frac{1}{10} > \frac{1}{31} + \frac{1}{32} + \frac{1}{33}$

$\Rightarrow \frac{1}{n} = \frac{1}{10} \Rightarrow n = 10$

and

$O R \frac{1}{31} > \frac{1}{33}; \frac{1}{32} > \frac{1}{33}; \frac{1}{33} = \frac{1}{33}$

$\Rightarrow \frac{1}{31} + \frac{1}{32} + \frac{1}{33} > \frac{3}{33}$

$\Rightarrow \frac{1}{31} + \frac{1}{32} + \frac{1}{33} > \frac{1}{11} = \frac{1}{n + 1}$

n = 10

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If1n+1<131+132+133<1n; then n?

The correct option is B 10130>133;130>131;130>131; ⇒130+130+130>131+132+133 ⇒330>131+132+133 ⇒110>131+132+133 ⇒1n=110⇒n=10 and OR 131>133;132>133;133=133 ⇒131+132+133>333 ⇒131+132+133>111=1n+1 n = 10

$I f \frac{1}{n + 1} < \frac{1}{31} + \frac{1}{32} + \frac{1}{33} < \frac{1}{n}; t h e n n ?$