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Question Number 151357 by mathdanisur last updated on 20/Aug/21

Commented by mr W last updated on 20/Aug/21

$c h e c k t h e q u e s t i o n p l e a s e!$ $\underset{k = 1}{\sum^{10}} a_{k} ⩾ 100 a n d \underset{k = 1}{\sum^{10}} a_{k}^{2} ⩽ 1001 ? ? ?$
Commented by mathdanisur last updated on 20/Aug/21

$Sorry Ser yes \underset{k = 1}{\sum^{10}} a_{k}^{2} ⩽ 1001$
Commented by mr W last updated on 20/Aug/21

$m i n a_{k} = 10 - \frac{3}{\sqrt{10}}$ $m a x a_{k} = 10 + \frac{3}{\sqrt{10}}$
Commented by mathdanisur last updated on 20/Aug/21

$Thank You Dear S er$ $if possible pkease$
Commented by mr W last updated on 20/Aug/21

${w h a t}^{'} s t h e a n s w e r i n y o u r b o o k ?$
Commented by mathdanisur last updated on 20/Aug/21

$Yes Dear S er, the answer is corrcet$ $solution please S er$
	Answered by mr W last updated on 20/Aug/21

	$t h e m e a n o f t h e 10 n u m b e r s i s 10 .$ $s a y t h e s m a l l e s t n u m b e r i s 10 - a,$ $s u c h t h a t t h i s n u m b e r i s a s s m a l l$ $a s p o s s i b l e, t h e o t h e r n i n e n u m b e r s$ $s h o u l d b e b i g g e r t h a n 10 a n d a s b i g$ $a s p o s s i b l e . s u c h t h a t Σ a_{k}^{2} i s a s s m a l l$ $a s p o s s i b l e, t h e s e n i n e n u m b e r s$ $s h o u l d b e a s s m a l l a s p o s s i b l e . t h i s$ $c a n o n l y b e r e a c h e d i f t h e n i n e$ $n u m b e r s a r e e q u a l . s o w e h a v e o n e$ $n u m b e r 10 - a, t h e o t h e r n i n e o n e s$ $a r e l a r g e r t h a n o r e q u a l t o 10 + \frac{a}{9} .$ $S = Σ a_{k}^{2} ⩾ {(10 - a)}^{2} + 9 \times {(10 + \frac{a}{9})}^{2}$ $s i n c e S ⩽ 1001,$ $1001 ⩾ {(10 - a)}^{2} + 9 \times {(10 + \frac{a}{9})}^{2}$ $\Rightarrow 9 ⩾ 10 a^{2}$ $\Rightarrow a ⩽ \frac{3}{\sqrt{10}}$ $t h e s m a l l e s t n u m b e r i s$ $10 - a ⩾ 10 - \frac{3}{\sqrt{10}}$ $t h a t m e a n s t h e s m a l l e s t n u m b e r$ $i s 10 - \frac{3}{\sqrt{10}} .$ $s i m i l a r l y, s a y t h e l a r g e s t n u m b e r$ $i s 10 + b, t h e n t h e o t h e r n i n e n u m b e r s$ $s h o u l d b e e q u a l a n d e q u a l t o 10 - \frac{b}{9} .$ $1001 ⩾ S = Σ a_{k}^{2} ⩾ {(10 + b)}^{2} + 9 \times {(10 - \frac{a}{9})}^{2}$ $\Rightarrow 9 ⩾ 10 b^{2}$ $\Rightarrow b ⩽ \frac{3}{\sqrt{10}}$ $t h e l a r g e s t n u m b e r i s$ $10 + b ⩽ 10 + \frac{3}{\sqrt{10}}$ $t h a t m e a n s t h e l a r g e s t n u m b e r$ $i s 10 + \frac{3}{\sqrt{10}} .$
	Commented by mathdanisur last updated on 20/Aug/21

	$Nice S er, Thank You$
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