Sequences and series(Arithmetic and Geometric progression)

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Anyone can help me solve this question？



The first three of four integers are in an a.p. and the last three are in g.p. Find these four numbers, given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.



Thanks!




arithmetic-progressions




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  • Try to show us your trying.
    – Michael Rozenberg
    21 hours ago










  • You ask if "anyone can help me solve this question?" so clearly you want clues not the actual answer.... use algebra... if the first integer is $a$ and the common difference is $d$ you can write the first three terms of the AP.... does that get you started?
    – Bruce
    21 hours ago














up vote
0
down vote

favorite












Anyone can help me solve this question？



The first three of four integers are in an a.p. and the last three are in g.p. Find these four numbers, given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.



Thanks!




arithmetic-progressions




share|cite|improve this question





















  • Try to show us your trying.
    – Michael Rozenberg
    21 hours ago










  • You ask if "anyone can help me solve this question?" so clearly you want clues not the actual answer.... use algebra... if the first integer is $a$ and the common difference is $d$ you can write the first three terms of the AP.... does that get you started?
    – Bruce
    21 hours ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Anyone can help me solve this question？



The first three of four integers are in an a.p. and the last three are in g.p. Find these four numbers, given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.



Thanks!




arithmetic-progressions




share|cite|improve this question













Anyone can help me solve this question？



The first three of four integers are in an a.p. and the last three are in g.p. Find these four numbers, given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.



Thanks!





arithmetic-progressions





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edited 21 hours ago









Javi

2,1381525




2,1381525









asked 21 hours ago









Lynn

1




1











  • Try to show us your trying.
    – Michael Rozenberg
    21 hours ago










  • You ask if "anyone can help me solve this question?" so clearly you want clues not the actual answer.... use algebra... if the first integer is $a$ and the common difference is $d$ you can write the first three terms of the AP.... does that get you started?
    – Bruce
    21 hours ago
















  • Try to show us your trying.
    – Michael Rozenberg
    21 hours ago










  • You ask if "anyone can help me solve this question?" so clearly you want clues not the actual answer.... use algebra... if the first integer is $a$ and the common difference is $d$ you can write the first three terms of the AP.... does that get you started?
    – Bruce
    21 hours ago















Try to show us your trying.
– Michael Rozenberg
21 hours ago




Try to show us your trying.
– Michael Rozenberg
21 hours ago












You ask if "anyone can help me solve this question?" so clearly you want clues not the actual answer.... use algebra... if the first integer is $a$ and the common difference is $d$ you can write the first three terms of the AP.... does that get you started?
– Bruce
21 hours ago




You ask if "anyone can help me solve this question?" so clearly you want clues not the actual answer.... use algebra... if the first integer is $a$ and the common difference is $d$ you can write the first three terms of the AP.... does that get you started?
– Bruce
21 hours ago










2 Answers
2






active

oldest

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up vote
1
down vote













Let the second digit be $n$, and the common difference of the AP be $d$, the common ratio of the GP be $r$:



$(n-d), n, nr, nr^2$



So as the middle two integers are also in AP we have that



$d=nr-n=n(r-1)$



so we can eliminate $d$ and have the four integers as



$(n-n(r-1)), n, nr, nr^2$



or more tidily



$n(2-r), n, nr, nr^2$



Now we can make a few equations based on the sums of the first and last etc that you state:



$n(2-r)+nr^2=37$



factorising out the $n$ gives



$n(r^2-r+2)=37$ ............... Equation 1



and the middle number sum:



$n+nr=36$



which tidies to



$n(1+r)=36$ ............... Equation 2



combining equations 1 and 2 to eliminate $n$ gives:



$frac361+r=frac37r^2-r+2$



which, I assure you, tidies to



$36r^2-73r+35=0$



solving the quadratic gives $r=frac54$. The other value it gives won't give integer solutions (try it yourself and see). Putting this value of $r$ into equation 2 gives us $n=16$. So we have



12, 16, 20, 25






share|cite|improve this answer




























    up vote
    1
    down vote













    Call the three numbers $a,b,c,d$. Then you know that $a+c=2b$ and $bd=c^2$. Next you know that $a+d=37$ and $b+c=36$.



    Leave aside the degree two equation. The linear ones can be solved for $b$ by $d=37-a$, $c=36-b$ and so
    $$
    a-2b+36-b=0
    $$
    that yields $a=3b-36$ and therefore $d=37-3b+36=73-3b$. Finally
    $$
    b(73-3b)=(36-b)^2
    $$
    Can you finish?






    share|cite|improve this answer





















    • +1..............To the proposer: (I). $b+c= 36 $ so we can express $c$ in terms of $b$....(II). $ a+c=2b$ so we can express $a$ in terms of $b$ and $c,$ but $c$ is expressible in in terms of $b$ by (I) so we can express $a$ in terms of $b$ only.... (III). $a+d=37$ so we can express $d$ in terms of $a,$ but by (II), $a$ is expressible in terms of $b$ only, so $d$ can be expressed in terms of $b$ only....(IV). Finally, by (I),(II), and (III), all the letters in $bd=c^2$ are expressible in terms of $b$ only. ...Once we find $b$ we find, successively, $ c=36-b, a=2b-c, d=37-a.$
      – DanielWainfleet
      8 hours ago











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    2 Answers
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    2 Answers
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    up vote
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    Let the second digit be $n$, and the common difference of the AP be $d$, the common ratio of the GP be $r$:



    $(n-d), n, nr, nr^2$



    So as the middle two integers are also in AP we have that



    $d=nr-n=n(r-1)$



    so we can eliminate $d$ and have the four integers as



    $(n-n(r-1)), n, nr, nr^2$



    or more tidily



    $n(2-r), n, nr, nr^2$



    Now we can make a few equations based on the sums of the first and last etc that you state:



    $n(2-r)+nr^2=37$



    factorising out the $n$ gives



    $n(r^2-r+2)=37$ ............... Equation 1



    and the middle number sum:



    $n+nr=36$



    which tidies to



    $n(1+r)=36$ ............... Equation 2



    combining equations 1 and 2 to eliminate $n$ gives:



    $frac361+r=frac37r^2-r+2$



    which, I assure you, tidies to



    $36r^2-73r+35=0$



    solving the quadratic gives $r=frac54$. The other value it gives won't give integer solutions (try it yourself and see). Putting this value of $r$ into equation 2 gives us $n=16$. So we have



    12, 16, 20, 25






    share|cite|improve this answer

























      up vote
      1
      down vote













      Let the second digit be $n$, and the common difference of the AP be $d$, the common ratio of the GP be $r$:



      $(n-d), n, nr, nr^2$



      So as the middle two integers are also in AP we have that



      $d=nr-n=n(r-1)$



      so we can eliminate $d$ and have the four integers as



      $(n-n(r-1)), n, nr, nr^2$



      or more tidily



      $n(2-r), n, nr, nr^2$



      Now we can make a few equations based on the sums of the first and last etc that you state:



      $n(2-r)+nr^2=37$



      factorising out the $n$ gives



      $n(r^2-r+2)=37$ ............... Equation 1



      and the middle number sum:



      $n+nr=36$



      which tidies to



      $n(1+r)=36$ ............... Equation 2



      combining equations 1 and 2 to eliminate $n$ gives:



      $frac361+r=frac37r^2-r+2$



      which, I assure you, tidies to



      $36r^2-73r+35=0$



      solving the quadratic gives $r=frac54$. The other value it gives won't give integer solutions (try it yourself and see). Putting this value of $r$ into equation 2 gives us $n=16$. So we have



      12, 16, 20, 25






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Let the second digit be $n$, and the common difference of the AP be $d$, the common ratio of the GP be $r$:



        $(n-d), n, nr, nr^2$



        So as the middle two integers are also in AP we have that



        $d=nr-n=n(r-1)$



        so we can eliminate $d$ and have the four integers as



        $(n-n(r-1)), n, nr, nr^2$



        or more tidily



        $n(2-r), n, nr, nr^2$



        Now we can make a few equations based on the sums of the first and last etc that you state:



        $n(2-r)+nr^2=37$



        factorising out the $n$ gives



        $n(r^2-r+2)=37$ ............... Equation 1



        and the middle number sum:



        $n+nr=36$



        which tidies to



        $n(1+r)=36$ ............... Equation 2



        combining equations 1 and 2 to eliminate $n$ gives:



        $frac361+r=frac37r^2-r+2$



        which, I assure you, tidies to



        $36r^2-73r+35=0$



        solving the quadratic gives $r=frac54$. The other value it gives won't give integer solutions (try it yourself and see). Putting this value of $r$ into equation 2 gives us $n=16$. So we have



        12, 16, 20, 25






        share|cite|improve this answer













        Let the second digit be $n$, and the common difference of the AP be $d$, the common ratio of the GP be $r$:



        $(n-d), n, nr, nr^2$



        So as the middle two integers are also in AP we have that



        $d=nr-n=n(r-1)$



        so we can eliminate $d$ and have the four integers as



        $(n-n(r-1)), n, nr, nr^2$



        or more tidily



        $n(2-r), n, nr, nr^2$



        Now we can make a few equations based on the sums of the first and last etc that you state:



        $n(2-r)+nr^2=37$



        factorising out the $n$ gives



        $n(r^2-r+2)=37$ ............... Equation 1



        and the middle number sum:



        $n+nr=36$



        which tidies to



        $n(1+r)=36$ ............... Equation 2



        combining equations 1 and 2 to eliminate $n$ gives:



        $frac361+r=frac37r^2-r+2$



        which, I assure you, tidies to



        $36r^2-73r+35=0$



        solving the quadratic gives $r=frac54$. The other value it gives won't give integer solutions (try it yourself and see). Putting this value of $r$ into equation 2 gives us $n=16$. So we have



        12, 16, 20, 25







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 13 hours ago









        Bruce

        13611




        13611




















            up vote
            1
            down vote













            Call the three numbers $a,b,c,d$. Then you know that $a+c=2b$ and $bd=c^2$. Next you know that $a+d=37$ and $b+c=36$.



            Leave aside the degree two equation. The linear ones can be solved for $b$ by $d=37-a$, $c=36-b$ and so
            $$
            a-2b+36-b=0
            $$
            that yields $a=3b-36$ and therefore $d=37-3b+36=73-3b$. Finally
            $$
            b(73-3b)=(36-b)^2
            $$
            Can you finish?






            share|cite|improve this answer





















            • +1..............To the proposer: (I). $b+c= 36 $ so we can express $c$ in terms of $b$....(II). $ a+c=2b$ so we can express $a$ in terms of $b$ and $c,$ but $c$ is expressible in in terms of $b$ by (I) so we can express $a$ in terms of $b$ only.... (III). $a+d=37$ so we can express $d$ in terms of $a,$ but by (II), $a$ is expressible in terms of $b$ only, so $d$ can be expressed in terms of $b$ only....(IV). Finally, by (I),(II), and (III), all the letters in $bd=c^2$ are expressible in terms of $b$ only. ...Once we find $b$ we find, successively, $ c=36-b, a=2b-c, d=37-a.$
              – DanielWainfleet
              8 hours ago















            up vote
            1
            down vote













            Call the three numbers $a,b,c,d$. Then you know that $a+c=2b$ and $bd=c^2$. Next you know that $a+d=37$ and $b+c=36$.



            Leave aside the degree two equation. The linear ones can be solved for $b$ by $d=37-a$, $c=36-b$ and so
            $$
            a-2b+36-b=0
            $$
            that yields $a=3b-36$ and therefore $d=37-3b+36=73-3b$. Finally
            $$
            b(73-3b)=(36-b)^2
            $$
            Can you finish?






            share|cite|improve this answer





















            • +1..............To the proposer: (I). $b+c= 36 $ so we can express $c$ in terms of $b$....(II). $ a+c=2b$ so we can express $a$ in terms of $b$ and $c,$ but $c$ is expressible in in terms of $b$ by (I) so we can express $a$ in terms of $b$ only.... (III). $a+d=37$ so we can express $d$ in terms of $a,$ but by (II), $a$ is expressible in terms of $b$ only, so $d$ can be expressed in terms of $b$ only....(IV). Finally, by (I),(II), and (III), all the letters in $bd=c^2$ are expressible in terms of $b$ only. ...Once we find $b$ we find, successively, $ c=36-b, a=2b-c, d=37-a.$
              – DanielWainfleet
              8 hours ago













            up vote
            1
            down vote










            up vote
            1
            down vote









            Call the three numbers $a,b,c,d$. Then you know that $a+c=2b$ and $bd=c^2$. Next you know that $a+d=37$ and $b+c=36$.



            Leave aside the degree two equation. The linear ones can be solved for $b$ by $d=37-a$, $c=36-b$ and so
            $$
            a-2b+36-b=0
            $$
            that yields $a=3b-36$ and therefore $d=37-3b+36=73-3b$. Finally
            $$
            b(73-3b)=(36-b)^2
            $$
            Can you finish?






            share|cite|improve this answer













            Call the three numbers $a,b,c,d$. Then you know that $a+c=2b$ and $bd=c^2$. Next you know that $a+d=37$ and $b+c=36$.



            Leave aside the degree two equation. The linear ones can be solved for $b$ by $d=37-a$, $c=36-b$ and so
            $$
            a-2b+36-b=0
            $$
            that yields $a=3b-36$ and therefore $d=37-3b+36=73-3b$. Finally
            $$
            b(73-3b)=(36-b)^2
            $$
            Can you finish?







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered 12 hours ago









            egreg

            164k1179187




            164k1179187











            • +1..............To the proposer: (I). $b+c= 36 $ so we can express $c$ in terms of $b$....(II). $ a+c=2b$ so we can express $a$ in terms of $b$ and $c,$ but $c$ is expressible in in terms of $b$ by (I) so we can express $a$ in terms of $b$ only.... (III). $a+d=37$ so we can express $d$ in terms of $a,$ but by (II), $a$ is expressible in terms of $b$ only, so $d$ can be expressed in terms of $b$ only....(IV). Finally, by (I),(II), and (III), all the letters in $bd=c^2$ are expressible in terms of $b$ only. ...Once we find $b$ we find, successively, $ c=36-b, a=2b-c, d=37-a.$
              – DanielWainfleet
              8 hours ago

















            • +1..............To the proposer: (I). $b+c= 36 $ so we can express $c$ in terms of $b$....(II). $ a+c=2b$ so we can express $a$ in terms of $b$ and $c,$ but $c$ is expressible in in terms of $b$ by (I) so we can express $a$ in terms of $b$ only.... (III). $a+d=37$ so we can express $d$ in terms of $a,$ but by (II), $a$ is expressible in terms of $b$ only, so $d$ can be expressed in terms of $b$ only....(IV). Finally, by (I),(II), and (III), all the letters in $bd=c^2$ are expressible in terms of $b$ only. ...Once we find $b$ we find, successively, $ c=36-b, a=2b-c, d=37-a.$
              – DanielWainfleet
              8 hours ago
















            +1..............To the proposer: (I). $b+c= 36 $ so we can express $c$ in terms of $b$....(II). $ a+c=2b$ so we can express $a$ in terms of $b$ and $c,$ but $c$ is expressible in in terms of $b$ by (I) so we can express $a$ in terms of $b$ only.... (III). $a+d=37$ so we can express $d$ in terms of $a,$ but by (II), $a$ is expressible in terms of $b$ only, so $d$ can be expressed in terms of $b$ only....(IV). Finally, by (I),(II), and (III), all the letters in $bd=c^2$ are expressible in terms of $b$ only. ...Once we find $b$ we find, successively, $ c=36-b, a=2b-c, d=37-a.$
            – DanielWainfleet
            8 hours ago





            +1..............To the proposer: (I). $b+c= 36 $ so we can express $c$ in terms of $b$....(II). $ a+c=2b$ so we can express $a$ in terms of $b$ and $c,$ but $c$ is expressible in in terms of $b$ by (I) so we can express $a$ in terms of $b$ only.... (III). $a+d=37$ so we can express $d$ in terms of $a,$ but by (II), $a$ is expressible in terms of $b$ only, so $d$ can be expressed in terms of $b$ only....(IV). Finally, by (I),(II), and (III), all the letters in $bd=c^2$ are expressible in terms of $b$ only. ...Once we find $b$ we find, successively, $ c=36-b, a=2b-c, d=37-a.$
            – DanielWainfleet
            8 hours ago













             

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