Find the sum of these variables.
$begingroup$
Five real numbers $a_1, a_2, a_3, a_4;text{and}; a_5;$ are such that
$$sqrt{a_1- 1} + 2sqrt{a_2- 4}+3sqrt{a_3- 9}
+4sqrt{a_4- 16} + 5 sqrt{a_4- 25} =frac{a_1+a_2+a_3+a_4+a_5}{2}.$$
Find $a_1+a_2+a_3+a_4+a_5.$
Thanks for checking this out!
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
Five real numbers $a_1, a_2, a_3, a_4;text{and}; a_5;$ are such that
$$sqrt{a_1- 1} + 2sqrt{a_2- 4}+3sqrt{a_3- 9}
+4sqrt{a_4- 16} + 5 sqrt{a_4- 25} =frac{a_1+a_2+a_3+a_4+a_5}{2}.$$
Find $a_1+a_2+a_3+a_4+a_5.$
Thanks for checking this out!
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
Five real numbers $a_1, a_2, a_3, a_4;text{and}; a_5;$ are such that
$$sqrt{a_1- 1} + 2sqrt{a_2- 4}+3sqrt{a_3- 9}
+4sqrt{a_4- 16} + 5 sqrt{a_4- 25} =frac{a_1+a_2+a_3+a_4+a_5}{2}.$$
Find $a_1+a_2+a_3+a_4+a_5.$
Thanks for checking this out!
algebra-precalculus
$endgroup$
Five real numbers $a_1, a_2, a_3, a_4;text{and}; a_5;$ are such that
$$sqrt{a_1- 1} + 2sqrt{a_2- 4}+3sqrt{a_3- 9}
+4sqrt{a_4- 16} + 5 sqrt{a_4- 25} =frac{a_1+a_2+a_3+a_4+a_5}{2}.$$
Find $a_1+a_2+a_3+a_4+a_5.$
Thanks for checking this out!
algebra-precalculus
algebra-precalculus
edited Dec 14 '18 at 19:07
user376343
3,7883828
3,7883828
asked Dec 14 '18 at 17:05
Sidharth A SSidharth A S
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For real $sqrt{a-b^2},$ we need $a-b^2ge0$
and for $b>0,$ and as $sqrt{a-b^2}ge0$ by AM-GM inequality,
$$dfrac{(sqrt{a-b^2})^2+(b)^2}2ge bsqrt{a-b^2}$$
the equality will occur if $sqrt{a-b^2}=b$
$$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$
$endgroup$
$begingroup$
@MathLover, Please free to rectify
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:18
$begingroup$
Sorry, but what does that first equation mean?
$endgroup$
– Sidharth A S
Dec 14 '18 at 17:19
$begingroup$
@SidharthAS, en.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:22
add a comment |
$begingroup$
Define$$b_1=sqrt{a_1-1}\b_2=sqrt{a_2-4}\b_3=sqrt{a_3-9}\b_4=sqrt{a_4-16}\b_5=sqrt{a_5-25}\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2over 2}+{55over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b_2-2)^2+(b_3-3)^2+(b_4-4)^2+(b_5-5)^2=0$$therefore $$b_i=iquad ,quad i=1,2,3,4,5$$and we have $$a_i=2i^2$$which yields to $$sum a_i=110$$
$endgroup$
$begingroup$
excellent solution!!!
$endgroup$
– user376343
Dec 16 '18 at 10:48
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For real $sqrt{a-b^2},$ we need $a-b^2ge0$
and for $b>0,$ and as $sqrt{a-b^2}ge0$ by AM-GM inequality,
$$dfrac{(sqrt{a-b^2})^2+(b)^2}2ge bsqrt{a-b^2}$$
the equality will occur if $sqrt{a-b^2}=b$
$$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$
$endgroup$
$begingroup$
@MathLover, Please free to rectify
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:18
$begingroup$
Sorry, but what does that first equation mean?
$endgroup$
– Sidharth A S
Dec 14 '18 at 17:19
$begingroup$
@SidharthAS, en.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:22
add a comment |
$begingroup$
For real $sqrt{a-b^2},$ we need $a-b^2ge0$
and for $b>0,$ and as $sqrt{a-b^2}ge0$ by AM-GM inequality,
$$dfrac{(sqrt{a-b^2})^2+(b)^2}2ge bsqrt{a-b^2}$$
the equality will occur if $sqrt{a-b^2}=b$
$$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$
$endgroup$
$begingroup$
@MathLover, Please free to rectify
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:18
$begingroup$
Sorry, but what does that first equation mean?
$endgroup$
– Sidharth A S
Dec 14 '18 at 17:19
$begingroup$
@SidharthAS, en.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:22
add a comment |
$begingroup$
For real $sqrt{a-b^2},$ we need $a-b^2ge0$
and for $b>0,$ and as $sqrt{a-b^2}ge0$ by AM-GM inequality,
$$dfrac{(sqrt{a-b^2})^2+(b)^2}2ge bsqrt{a-b^2}$$
the equality will occur if $sqrt{a-b^2}=b$
$$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$
$endgroup$
For real $sqrt{a-b^2},$ we need $a-b^2ge0$
and for $b>0,$ and as $sqrt{a-b^2}ge0$ by AM-GM inequality,
$$dfrac{(sqrt{a-b^2})^2+(b)^2}2ge bsqrt{a-b^2}$$
the equality will occur if $sqrt{a-b^2}=b$
$$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$
edited Dec 14 '18 at 17:39
answered Dec 14 '18 at 17:16
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
@MathLover, Please free to rectify
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:18
$begingroup$
Sorry, but what does that first equation mean?
$endgroup$
– Sidharth A S
Dec 14 '18 at 17:19
$begingroup$
@SidharthAS, en.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:22
add a comment |
$begingroup$
@MathLover, Please free to rectify
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:18
$begingroup$
Sorry, but what does that first equation mean?
$endgroup$
– Sidharth A S
Dec 14 '18 at 17:19
$begingroup$
@SidharthAS, en.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:22
$begingroup$
@MathLover, Please free to rectify
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:18
$begingroup$
@MathLover, Please free to rectify
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:18
$begingroup$
Sorry, but what does that first equation mean?
$endgroup$
– Sidharth A S
Dec 14 '18 at 17:19
$begingroup$
Sorry, but what does that first equation mean?
$endgroup$
– Sidharth A S
Dec 14 '18 at 17:19
$begingroup$
@SidharthAS, en.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:22
$begingroup$
@SidharthAS, en.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 17:22
add a comment |
$begingroup$
Define$$b_1=sqrt{a_1-1}\b_2=sqrt{a_2-4}\b_3=sqrt{a_3-9}\b_4=sqrt{a_4-16}\b_5=sqrt{a_5-25}\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2over 2}+{55over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b_2-2)^2+(b_3-3)^2+(b_4-4)^2+(b_5-5)^2=0$$therefore $$b_i=iquad ,quad i=1,2,3,4,5$$and we have $$a_i=2i^2$$which yields to $$sum a_i=110$$
$endgroup$
$begingroup$
excellent solution!!!
$endgroup$
– user376343
Dec 16 '18 at 10:48
add a comment |
$begingroup$
Define$$b_1=sqrt{a_1-1}\b_2=sqrt{a_2-4}\b_3=sqrt{a_3-9}\b_4=sqrt{a_4-16}\b_5=sqrt{a_5-25}\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2over 2}+{55over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b_2-2)^2+(b_3-3)^2+(b_4-4)^2+(b_5-5)^2=0$$therefore $$b_i=iquad ,quad i=1,2,3,4,5$$and we have $$a_i=2i^2$$which yields to $$sum a_i=110$$
$endgroup$
$begingroup$
excellent solution!!!
$endgroup$
– user376343
Dec 16 '18 at 10:48
add a comment |
$begingroup$
Define$$b_1=sqrt{a_1-1}\b_2=sqrt{a_2-4}\b_3=sqrt{a_3-9}\b_4=sqrt{a_4-16}\b_5=sqrt{a_5-25}\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2over 2}+{55over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b_2-2)^2+(b_3-3)^2+(b_4-4)^2+(b_5-5)^2=0$$therefore $$b_i=iquad ,quad i=1,2,3,4,5$$and we have $$a_i=2i^2$$which yields to $$sum a_i=110$$
$endgroup$
Define$$b_1=sqrt{a_1-1}\b_2=sqrt{a_2-4}\b_3=sqrt{a_3-9}\b_4=sqrt{a_4-16}\b_5=sqrt{a_5-25}\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2over 2}+{55over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b_2-2)^2+(b_3-3)^2+(b_4-4)^2+(b_5-5)^2=0$$therefore $$b_i=iquad ,quad i=1,2,3,4,5$$and we have $$a_i=2i^2$$which yields to $$sum a_i=110$$
answered Dec 14 '18 at 21:25
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
excellent solution!!!
$endgroup$
– user376343
Dec 16 '18 at 10:48
add a comment |
$begingroup$
excellent solution!!!
$endgroup$
– user376343
Dec 16 '18 at 10:48
$begingroup$
excellent solution!!!
$endgroup$
– user376343
Dec 16 '18 at 10:48
$begingroup$
excellent solution!!!
$endgroup$
– user376343
Dec 16 '18 at 10:48
add a comment |
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