Find the sum of these variables.












1












$begingroup$


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







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$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 19:07









      user376343

      3,7783827




      3,7783827










      asked Dec 14 '18 at 17:05









      Sidharth A SSidharth A S

      111




      111






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          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 $sqrt{a-b^2}=b$



          $$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @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



















          2












          $begingroup$

          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$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            excellent solution!!!
            $endgroup$
            – user376343
            Dec 16 '18 at 10:48











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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 $sqrt{a-b^2}=b$



          $$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @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
















          4












          $begingroup$

          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 $sqrt{a-b^2}=b$



          $$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @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














          4












          4








          4





          $begingroup$

          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 $sqrt{a-b^2}=b$



          $$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$






          share|cite|improve this answer











          $endgroup$



          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 $sqrt{a-b^2}=b$



          $$impliesdfrac{(sqrt{a_1-1})^2+1^2+cdots+(sqrt{a_5-5^2})^2+5^2}2=dfrac{a_1+a_2+a_3+a_4+a_5}2 ge sqrt{a_1-1}+cdots$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 17:39

























          answered Dec 14 '18 at 17:16









          lab bhattacharjeelab bhattacharjee

          225k15157275




          225k15157275












          • $begingroup$
            @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










          • $begingroup$
            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











          2












          $begingroup$

          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$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            excellent solution!!!
            $endgroup$
            – user376343
            Dec 16 '18 at 10:48
















          2












          $begingroup$

          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$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            excellent solution!!!
            $endgroup$
            – user376343
            Dec 16 '18 at 10:48














          2












          2








          2





          $begingroup$

          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$$






          share|cite|improve this answer









          $endgroup$



          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$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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