Sum of squares in geometric progression












5












$begingroup$



In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






algebra-precalculus geometric-progressions







share|cite|improve this question











$endgroup$

















    5












    $begingroup$



    In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
    and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




    If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






    algebra-precalculus geometric-progressions







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$



      In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
      and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




      If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






      algebra-precalculus geometric-progressions







      share|cite|improve this question











      $endgroup$





      In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
      and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




      If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






      algebra-precalculus geometric-progressions




      algebra-precalculus geometric-progressions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 2 '15 at 6:18









      nbubis

      27.4k553110




      27.4k553110










      asked Mar 2 '15 at 4:44







      user167045





























          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27



















          2












          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11












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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27
















          6












          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27














          6












          6








          6





          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$



          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 2 '15 at 5:08

























          answered Mar 2 '15 at 4:56









          Janko BracicJanko Bracic

          2,783613




          2,783613












          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27


















          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27
















          $begingroup$
          After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
          $endgroup$
          – user167045
          Mar 2 '15 at 5:04






          $begingroup$
          After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
          $endgroup$
          – user167045
          Mar 2 '15 at 5:04














          $begingroup$
          @Garvil I don't see any leftover terms.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:10




          $begingroup$
          @Garvil I don't see any leftover terms.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:10












          $begingroup$
          Yeah, sorry. My bad.
          $endgroup$
          – user167045
          Mar 2 '15 at 5:12




          $begingroup$
          Yeah, sorry. My bad.
          $endgroup$
          – user167045
          Mar 2 '15 at 5:12












          $begingroup$
          No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:13




          $begingroup$
          No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:13












          $begingroup$
          Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
          $endgroup$
          – Janko Bracic
          Mar 2 '15 at 5:27




          $begingroup$
          Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
          $endgroup$
          – Janko Bracic
          Mar 2 '15 at 5:27











          2












          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11
















          2












          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11














          2












          2








          2





          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$



          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 30 '18 at 1:41









          Rócherz

          3,0263823




          3,0263823










          answered Mar 2 '15 at 5:08









          KugelblitzKugelblitz

          3,3491449




          3,3491449












          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11


















          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11
















          $begingroup$
          Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:09






          $begingroup$
          Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:09














          $begingroup$
          20 terms cancel out; remaining five are the required five.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:11




          $begingroup$
          20 terms cancel out; remaining five are the required five.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:11


















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