find the first term of the series?












0












$begingroup$


The sum of an infinite geometric series of real numbers is $14,$ and the
sum of the cubes of the terms of this series is $392$.



What is the first term of the series?



My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
$$ s = frac{a}{1 - r} = 14 tag{1}$$
When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
which sums to
$$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$



Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.



Any hints/solution is appreciated.






sequences-and-series algebra-precalculus summation geometric-progressions







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$endgroup$

















    0












    $begingroup$


    The sum of an infinite geometric series of real numbers is $14,$ and the
    sum of the cubes of the terms of this series is $392$.



    What is the first term of the series?



    My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
    $$ s = frac{a}{1 - r} = 14 tag{1}$$
    When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
    which sums to
    $$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$



    Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.



    Any hints/solution is appreciated.






    sequences-and-series algebra-precalculus summation geometric-progressions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The sum of an infinite geometric series of real numbers is $14,$ and the
      sum of the cubes of the terms of this series is $392$.



      What is the first term of the series?



      My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
      $$ s = frac{a}{1 - r} = 14 tag{1}$$
      When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
      which sums to
      $$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$



      Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.



      Any hints/solution is appreciated.






      sequences-and-series algebra-precalculus summation geometric-progressions







      share|cite|improve this question











      $endgroup$




      The sum of an infinite geometric series of real numbers is $14,$ and the
      sum of the cubes of the terms of this series is $392$.



      What is the first term of the series?



      My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
      $$ s = frac{a}{1 - r} = 14 tag{1}$$
      When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
      which sums to
      $$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$



      Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.



      Any hints/solution is appreciated.






      sequences-and-series algebra-precalculus summation geometric-progressions




      sequences-and-series algebra-precalculus summation geometric-progressions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 17:05









      Lee David Chung Lin

      3,86531140




      3,86531140










      asked Jan 1 at 21:14









      jasminejasmine

      1,650416




      1,650416






















          2 Answers
          2






          active

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          2












          $begingroup$

          So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
          $$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
          and so
          $$(2r-1)(r-2)=0implies r=1/2$$
          since $|r|<1.$ And so $a=7.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how $1+ r+r^2 = 1 +r^2-2r$? im not getting
            $endgroup$
            – jasmine
            Jan 1 at 21:26






          • 1




            $begingroup$
            They are not equal. $28(1+r+r^2) = 196(1-r)^2$
            $endgroup$
            – Hello_World
            Jan 1 at 21:30










          • $begingroup$
            okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
            $endgroup$
            – jasmine
            Jan 1 at 21:38





















          2












          $begingroup$

          You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
            $$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
            and so
            $$(2r-1)(r-2)=0implies r=1/2$$
            since $|r|<1.$ And so $a=7.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how $1+ r+r^2 = 1 +r^2-2r$? im not getting
              $endgroup$
              – jasmine
              Jan 1 at 21:26






            • 1




              $begingroup$
              They are not equal. $28(1+r+r^2) = 196(1-r)^2$
              $endgroup$
              – Hello_World
              Jan 1 at 21:30










            • $begingroup$
              okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
              $endgroup$
              – jasmine
              Jan 1 at 21:38


















            2












            $begingroup$

            So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
            $$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
            and so
            $$(2r-1)(r-2)=0implies r=1/2$$
            since $|r|<1.$ And so $a=7.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how $1+ r+r^2 = 1 +r^2-2r$? im not getting
              $endgroup$
              – jasmine
              Jan 1 at 21:26






            • 1




              $begingroup$
              They are not equal. $28(1+r+r^2) = 196(1-r)^2$
              $endgroup$
              – Hello_World
              Jan 1 at 21:30










            • $begingroup$
              okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
              $endgroup$
              – jasmine
              Jan 1 at 21:38
















            2












            2








            2





            $begingroup$

            So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
            $$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
            and so
            $$(2r-1)(r-2)=0implies r=1/2$$
            since $|r|<1.$ And so $a=7.$






            share|cite|improve this answer









            $endgroup$



            So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
            $$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
            and so
            $$(2r-1)(r-2)=0implies r=1/2$$
            since $|r|<1.$ And so $a=7.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 21:21









            Hello_WorldHello_World

            4,16221631




            4,16221631












            • $begingroup$
              how $1+ r+r^2 = 1 +r^2-2r$? im not getting
              $endgroup$
              – jasmine
              Jan 1 at 21:26






            • 1




              $begingroup$
              They are not equal. $28(1+r+r^2) = 196(1-r)^2$
              $endgroup$
              – Hello_World
              Jan 1 at 21:30










            • $begingroup$
              okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
              $endgroup$
              – jasmine
              Jan 1 at 21:38




















            • $begingroup$
              how $1+ r+r^2 = 1 +r^2-2r$? im not getting
              $endgroup$
              – jasmine
              Jan 1 at 21:26






            • 1




              $begingroup$
              They are not equal. $28(1+r+r^2) = 196(1-r)^2$
              $endgroup$
              – Hello_World
              Jan 1 at 21:30










            • $begingroup$
              okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
              $endgroup$
              – jasmine
              Jan 1 at 21:38


















            $begingroup$
            how $1+ r+r^2 = 1 +r^2-2r$? im not getting
            $endgroup$
            – jasmine
            Jan 1 at 21:26




            $begingroup$
            how $1+ r+r^2 = 1 +r^2-2r$? im not getting
            $endgroup$
            – jasmine
            Jan 1 at 21:26




            1




            1




            $begingroup$
            They are not equal. $28(1+r+r^2) = 196(1-r)^2$
            $endgroup$
            – Hello_World
            Jan 1 at 21:30




            $begingroup$
            They are not equal. $28(1+r+r^2) = 196(1-r)^2$
            $endgroup$
            – Hello_World
            Jan 1 at 21:30












            $begingroup$
            okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
            $endgroup$
            – jasmine
            Jan 1 at 21:38






            $begingroup$
            okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
            $endgroup$
            – jasmine
            Jan 1 at 21:38













            2












            $begingroup$

            You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.






                share|cite|improve this answer









                $endgroup$



                You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 1 at 21:23









                José Carlos SantosJosé Carlos Santos

                153k22123225




                153k22123225






























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