Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

check the question please! Σ_(k=1) ^(10) a_k ≥100 and Σ_(k=1) ^(10) a_k ^2 ≤1001 ???

$${check}\:{the}\:{question}\:{please}! \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{a}_{{k}} \geqslant\mathrm{100}\:{and}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{a}_{{k}} ^{\mathrm{2}} \leqslant\mathrm{1001}\:\:\:??? \\ $$

Commented by mathdanisur last updated on 20/Aug/21

Sorry Ser yes Σ_( k=1) ^(10) a_k ^2 ≤ 1001

$$\mathrm{Sorry}\:\mathrm{Ser}\:\mathrm{yes}\:\underset{\:\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} ^{\mathrm{2}} \:\leqslant\:\mathrm{1001} \\ $$

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

min a_k =10−(3/( (√(10)))) max a_k =10+(3/( (√(10))))

$${min}\:{a}_{{k}} =\mathrm{10}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${max}\:{a}_{{k}} =\mathrm{10}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$

Commented by mathdanisur last updated on 20/Aug/21

Thank You Dear Ser if possible pkease

$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$$$\mathrm{if}\:\mathrm{possible}\:\mathrm{pkease} \\ $$

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

what′s the answer in your book?

$${what}'{s}\:{the}\:{answer}\:{in}\:{your}\:{book}? \\ $$

Commented by mathdanisur last updated on 20/Aug/21

Yes Dear Ser, the answer is corrcet solution please Ser

$$\mathrm{Yes}\:\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{corrcet} \\ $$$$\mathrm{solution}\:\mathrm{please}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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

the mean of the 10 numbers is 10. say the smallest number is 10−a, such that this number is as small as possible, the other nine numbers should be bigger than 10 and as big as possible. such that Σa_k ^2 is as small as possible, these nine numbers should be as small as possible. this can only be reached if the nine numbers are equal. so we have one number 10−a, the other nine ones are larger than or equal to 10+(a/9). S=Σa_k ^2 ≥(10−a)^2 +9×(10+(a/9))^2 since S≤1001, 1001≥(10−a)^2 +9×(10+(a/9))^2 ⇒9≥10a^2 ⇒a≤(3/( (√(10)))) the smallest number is 10−a≥10−(3/( (√(10)))) that means the smallest number is 10−(3/( (√(10)))). similarly, say the largest number is 10+b, then the other nine numbers should be equal and equal to 10−(b/9). 1001≥S=Σa_k ^2 ≥(10+b)^2 +9×(10−(a/9))^2 ⇒9≥10b^2 ⇒b≤(3/( (√(10)))) the largest number is 10+b≤10+(3/( (√(10)))) that means the largest number is 10+(3/( (√(10)))).

$${the}\:{mean}\:{of}\:{the}\:\mathrm{10}\:{numbers}\:{is}\:\mathrm{10}. \\ $$$${say}\:{the}\:{smallest}\:{number}\:{is}\:\mathrm{10}−{a}, \\ $$$${such}\:{that}\:{this}\:{number}\:{is}\:{as}\:{small} \\ $$$${as}\:{possible},\:{the}\:{other}\:{nine}\:{numbers} \\ $$$${should}\:{be}\:{bigger}\:{than}\:\mathrm{10}\:{and}\:{as}\:{big} \\ $$$${as}\:{possible}.\:{such}\:{that}\:\Sigma{a}_{{k}} ^{\mathrm{2}} \:{is}\:{as}\:{small} \\ $$$${as}\:{possible},\:{these}\:{nine}\:{numbers}\: \\ $$$${should}\:{be}\:{as}\:{small}\:{as}\:{possible}.\:{this} \\ $$$${can}\:{only}\:{be}\:{reached}\:{if}\:{the}\:{nine} \\ $$$${numbers}\:{are}\:{equal}.\:{so}\:{we}\:{have}\:{one} \\ $$$${number}\:\mathrm{10}−{a},\:{the}\:{other}\:{nine}\:{ones} \\ $$$${are}\:{larger}\:{than}\:{or}\:{equal}\:{to}\:\mathrm{10}+\frac{{a}}{\mathrm{9}}. \\ $$$${S}=\Sigma{a}_{{k}} ^{\mathrm{2}} \geqslant\left(\mathrm{10}−{a}\right)^{\mathrm{2}} +\mathrm{9}×\left(\mathrm{10}+\frac{{a}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$${since}\:{S}\leqslant\mathrm{1001}, \\ $$$$\mathrm{1001}\geqslant\left(\mathrm{10}−{a}\right)^{\mathrm{2}} +\mathrm{9}×\left(\mathrm{10}+\frac{{a}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}\geqslant\mathrm{10}{a}^{\mathrm{2}} \\ $$$$\Rightarrow{a}\leqslant\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${the}\:{smallest}\:{number}\:{is} \\ $$$$\mathrm{10}−{a}\geqslant\mathrm{10}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${that}\:{means}\:{the}\:{smallest}\:{number} \\ $$$${is}\:\mathrm{10}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}. \\ $$$${similarly},\:{say}\:{the}\:{largest}\:{number} \\ $$$${is}\:\mathrm{10}+{b},\:{then}\:{the}\:{other}\:{nine}\:{numbers} \\ $$$${should}\:{be}\:{equal}\:{and}\:{equal}\:{to}\:\mathrm{10}−\frac{{b}}{\mathrm{9}}. \\ $$$$\mathrm{1001}\geqslant{S}=\Sigma{a}_{{k}} ^{\mathrm{2}} \geqslant\left(\mathrm{10}+{b}\right)^{\mathrm{2}} +\mathrm{9}×\left(\mathrm{10}−\frac{{a}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}\geqslant\mathrm{10}{b}^{\mathrm{2}} \\ $$$$\Rightarrow{b}\leqslant\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${the}\:{largest}\:{number}\:{is} \\ $$$$\mathrm{10}+{b}\leqslant\mathrm{10}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${that}\:{means}\:{the}\:{largest}\:{number} \\ $$$${is}\:\mathrm{10}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}. \\ $$

Commented by mathdanisur last updated on 20/Aug/21

Nice Ser, Thank You

$$\mathrm{Nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{Thank}\:\mathrm{You} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com