How to find the sum of the first 21 terms of an Arithmetic Progression(A.P)?












1












$begingroup$



Question. If the sum of first $12$ terms of an A.P. is equal to to the sum of the first $18$ terms of the same A.P., find the sum of the first $21$ terms of the same A.P.




$a=$ first term



$d=$ difference



I know now, $2(2a + 11d) = 3(2a + 17d)$



hence, $2a + 29d = 0$ --> correction



How do I proceed from here?






arithmetic-progressions







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Nick Using the formula ,Sum of n terms = n(2a + (n-1)d)/2
    $endgroup$
    – mac07
    Aug 13 '16 at 5:36










  • $begingroup$
    One possible solution is 0. Given the all zero sequence $x_n = 0$ (so $a=0$ and $d=0$), then we can verify that $sum_{i=1}^{12} x_i = sum_{i=1}^{12} 0 = 0 = sum_{i=1}^{18} 0 = sum_{i=1}^{18} x_i$. Since the sequence $x_n$ satisfies the necessary properties, we see that the sum of the first 21 terms is $sum_{i=1}^{21} x_i = sum_{i=1}^{21} 0 = 0$.
    $endgroup$
    – benguin
    Aug 13 '16 at 5:43












  • $begingroup$
    19th term+20th term+21st term$= 3a+57d$ and $S_{14}+3a+57d=21a+210d=S_{21}$ and $S_{18}+3a+57d=S_{21}$ to get 2 linear equations
    $endgroup$
    – Nick
    Aug 13 '16 at 5:44








  • 1




    $begingroup$
    Hint: get the second equation by equating the sum of $x_{13} to x_(18}$ to zero.
    $endgroup$
    – Moti
    Aug 13 '16 at 6:04








  • 1




    $begingroup$
    @benguin I finally got two equations : 6a + 87d = 0 and 2a + 27d = 0. Solve these equations give a=d=0
    $endgroup$
    – mac07
    Aug 13 '16 at 7:16
















1












$begingroup$



Question. If the sum of first $12$ terms of an A.P. is equal to to the sum of the first $18$ terms of the same A.P., find the sum of the first $21$ terms of the same A.P.




$a=$ first term



$d=$ difference



I know now, $2(2a + 11d) = 3(2a + 17d)$



hence, $2a + 29d = 0$ --> correction



How do I proceed from here?






arithmetic-progressions







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Nick Using the formula ,Sum of n terms = n(2a + (n-1)d)/2
    $endgroup$
    – mac07
    Aug 13 '16 at 5:36










  • $begingroup$
    One possible solution is 0. Given the all zero sequence $x_n = 0$ (so $a=0$ and $d=0$), then we can verify that $sum_{i=1}^{12} x_i = sum_{i=1}^{12} 0 = 0 = sum_{i=1}^{18} 0 = sum_{i=1}^{18} x_i$. Since the sequence $x_n$ satisfies the necessary properties, we see that the sum of the first 21 terms is $sum_{i=1}^{21} x_i = sum_{i=1}^{21} 0 = 0$.
    $endgroup$
    – benguin
    Aug 13 '16 at 5:43












  • $begingroup$
    19th term+20th term+21st term$= 3a+57d$ and $S_{14}+3a+57d=21a+210d=S_{21}$ and $S_{18}+3a+57d=S_{21}$ to get 2 linear equations
    $endgroup$
    – Nick
    Aug 13 '16 at 5:44








  • 1




    $begingroup$
    Hint: get the second equation by equating the sum of $x_{13} to x_(18}$ to zero.
    $endgroup$
    – Moti
    Aug 13 '16 at 6:04








  • 1




    $begingroup$
    @benguin I finally got two equations : 6a + 87d = 0 and 2a + 27d = 0. Solve these equations give a=d=0
    $endgroup$
    – mac07
    Aug 13 '16 at 7:16














1












1








1


1



$begingroup$



Question. If the sum of first $12$ terms of an A.P. is equal to to the sum of the first $18$ terms of the same A.P., find the sum of the first $21$ terms of the same A.P.




$a=$ first term



$d=$ difference



I know now, $2(2a + 11d) = 3(2a + 17d)$



hence, $2a + 29d = 0$ --> correction



How do I proceed from here?






arithmetic-progressions







share|cite|improve this question











$endgroup$





Question. If the sum of first $12$ terms of an A.P. is equal to to the sum of the first $18$ terms of the same A.P., find the sum of the first $21$ terms of the same A.P.




$a=$ first term



$d=$ difference



I know now, $2(2a + 11d) = 3(2a + 17d)$



hence, $2a + 29d = 0$ --> correction



How do I proceed from here?






arithmetic-progressions




arithmetic-progressions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 13 '16 at 5:38







mac07

















asked Aug 13 '16 at 5:22









mac07mac07

88212




88212












  • $begingroup$
    @Nick Using the formula ,Sum of n terms = n(2a + (n-1)d)/2
    $endgroup$
    – mac07
    Aug 13 '16 at 5:36










  • $begingroup$
    One possible solution is 0. Given the all zero sequence $x_n = 0$ (so $a=0$ and $d=0$), then we can verify that $sum_{i=1}^{12} x_i = sum_{i=1}^{12} 0 = 0 = sum_{i=1}^{18} 0 = sum_{i=1}^{18} x_i$. Since the sequence $x_n$ satisfies the necessary properties, we see that the sum of the first 21 terms is $sum_{i=1}^{21} x_i = sum_{i=1}^{21} 0 = 0$.
    $endgroup$
    – benguin
    Aug 13 '16 at 5:43












  • $begingroup$
    19th term+20th term+21st term$= 3a+57d$ and $S_{14}+3a+57d=21a+210d=S_{21}$ and $S_{18}+3a+57d=S_{21}$ to get 2 linear equations
    $endgroup$
    – Nick
    Aug 13 '16 at 5:44








  • 1




    $begingroup$
    Hint: get the second equation by equating the sum of $x_{13} to x_(18}$ to zero.
    $endgroup$
    – Moti
    Aug 13 '16 at 6:04








  • 1




    $begingroup$
    @benguin I finally got two equations : 6a + 87d = 0 and 2a + 27d = 0. Solve these equations give a=d=0
    $endgroup$
    – mac07
    Aug 13 '16 at 7:16


















  • $begingroup$
    @Nick Using the formula ,Sum of n terms = n(2a + (n-1)d)/2
    $endgroup$
    – mac07
    Aug 13 '16 at 5:36










  • $begingroup$
    One possible solution is 0. Given the all zero sequence $x_n = 0$ (so $a=0$ and $d=0$), then we can verify that $sum_{i=1}^{12} x_i = sum_{i=1}^{12} 0 = 0 = sum_{i=1}^{18} 0 = sum_{i=1}^{18} x_i$. Since the sequence $x_n$ satisfies the necessary properties, we see that the sum of the first 21 terms is $sum_{i=1}^{21} x_i = sum_{i=1}^{21} 0 = 0$.
    $endgroup$
    – benguin
    Aug 13 '16 at 5:43












  • $begingroup$
    19th term+20th term+21st term$= 3a+57d$ and $S_{14}+3a+57d=21a+210d=S_{21}$ and $S_{18}+3a+57d=S_{21}$ to get 2 linear equations
    $endgroup$
    – Nick
    Aug 13 '16 at 5:44








  • 1




    $begingroup$
    Hint: get the second equation by equating the sum of $x_{13} to x_(18}$ to zero.
    $endgroup$
    – Moti
    Aug 13 '16 at 6:04








  • 1




    $begingroup$
    @benguin I finally got two equations : 6a + 87d = 0 and 2a + 27d = 0. Solve these equations give a=d=0
    $endgroup$
    – mac07
    Aug 13 '16 at 7:16
















$begingroup$
@Nick Using the formula ,Sum of n terms = n(2a + (n-1)d)/2
$endgroup$
– mac07
Aug 13 '16 at 5:36




$begingroup$
@Nick Using the formula ,Sum of n terms = n(2a + (n-1)d)/2
$endgroup$
– mac07
Aug 13 '16 at 5:36












$begingroup$
One possible solution is 0. Given the all zero sequence $x_n = 0$ (so $a=0$ and $d=0$), then we can verify that $sum_{i=1}^{12} x_i = sum_{i=1}^{12} 0 = 0 = sum_{i=1}^{18} 0 = sum_{i=1}^{18} x_i$. Since the sequence $x_n$ satisfies the necessary properties, we see that the sum of the first 21 terms is $sum_{i=1}^{21} x_i = sum_{i=1}^{21} 0 = 0$.
$endgroup$
– benguin
Aug 13 '16 at 5:43






$begingroup$
One possible solution is 0. Given the all zero sequence $x_n = 0$ (so $a=0$ and $d=0$), then we can verify that $sum_{i=1}^{12} x_i = sum_{i=1}^{12} 0 = 0 = sum_{i=1}^{18} 0 = sum_{i=1}^{18} x_i$. Since the sequence $x_n$ satisfies the necessary properties, we see that the sum of the first 21 terms is $sum_{i=1}^{21} x_i = sum_{i=1}^{21} 0 = 0$.
$endgroup$
– benguin
Aug 13 '16 at 5:43














$begingroup$
19th term+20th term+21st term$= 3a+57d$ and $S_{14}+3a+57d=21a+210d=S_{21}$ and $S_{18}+3a+57d=S_{21}$ to get 2 linear equations
$endgroup$
– Nick
Aug 13 '16 at 5:44






$begingroup$
19th term+20th term+21st term$= 3a+57d$ and $S_{14}+3a+57d=21a+210d=S_{21}$ and $S_{18}+3a+57d=S_{21}$ to get 2 linear equations
$endgroup$
– Nick
Aug 13 '16 at 5:44






1




1




$begingroup$
Hint: get the second equation by equating the sum of $x_{13} to x_(18}$ to zero.
$endgroup$
– Moti
Aug 13 '16 at 6:04






$begingroup$
Hint: get the second equation by equating the sum of $x_{13} to x_(18}$ to zero.
$endgroup$
– Moti
Aug 13 '16 at 6:04






1




1




$begingroup$
@benguin I finally got two equations : 6a + 87d = 0 and 2a + 27d = 0. Solve these equations give a=d=0
$endgroup$
– mac07
Aug 13 '16 at 7:16




$begingroup$
@benguin I finally got two equations : 6a + 87d = 0 and 2a + 27d = 0. Solve these equations give a=d=0
$endgroup$
– mac07
Aug 13 '16 at 7:16










1 Answer
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$begingroup$

An arithmetic progression such that the sum of the first $12$ terms equals the sum of the first $18$ terms:
begin{multline}
(-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
(-17) + (-15) + (-13) + (-11) + (-9) + (-7) = -216,\
end{multline}
begin{multline}
(-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
(-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
(-5) + (-3) + (-1) + 1 + 3 + 5 = -216. \
end{multline}



The sum of the first $21$ terms of the series is



begin{multline}
(-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
(-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
(-5) + (-3) + (-1) + 1 + 3 + 5 +7 + 9 + 11 = -189. \
end{multline}



So it is possible to do this with non-zero $a$ and $d,$ namely
$a = -29,$ $d = 2.$



By the distributive law, we have equality of the sum of the first $12$ terms and the sum of the first $18$ terms whenever $a = -29k$ and $d = 2k$
for some constant $k.$
And then the sum of the first $21$ terms is $-189k.$



So the sum of the first $21$ terms could be any real number if $k$ is allowed to be any real number. It can be any multiple of $189$ if the terms must all be integers.



I suspect an error in the transcription of the question, possibly before it was presented to you.






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    $begingroup$

    An arithmetic progression such that the sum of the first $12$ terms equals the sum of the first $18$ terms:
    begin{multline}
    (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
    (-17) + (-15) + (-13) + (-11) + (-9) + (-7) = -216,\
    end{multline}
    begin{multline}
    (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
    (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
    (-5) + (-3) + (-1) + 1 + 3 + 5 = -216. \
    end{multline}



    The sum of the first $21$ terms of the series is



    begin{multline}
    (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
    (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
    (-5) + (-3) + (-1) + 1 + 3 + 5 +7 + 9 + 11 = -189. \
    end{multline}



    So it is possible to do this with non-zero $a$ and $d,$ namely
    $a = -29,$ $d = 2.$



    By the distributive law, we have equality of the sum of the first $12$ terms and the sum of the first $18$ terms whenever $a = -29k$ and $d = 2k$
    for some constant $k.$
    And then the sum of the first $21$ terms is $-189k.$



    So the sum of the first $21$ terms could be any real number if $k$ is allowed to be any real number. It can be any multiple of $189$ if the terms must all be integers.



    I suspect an error in the transcription of the question, possibly before it was presented to you.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      An arithmetic progression such that the sum of the first $12$ terms equals the sum of the first $18$ terms:
      begin{multline}
      (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
      (-17) + (-15) + (-13) + (-11) + (-9) + (-7) = -216,\
      end{multline}
      begin{multline}
      (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
      (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
      (-5) + (-3) + (-1) + 1 + 3 + 5 = -216. \
      end{multline}



      The sum of the first $21$ terms of the series is



      begin{multline}
      (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
      (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
      (-5) + (-3) + (-1) + 1 + 3 + 5 +7 + 9 + 11 = -189. \
      end{multline}



      So it is possible to do this with non-zero $a$ and $d,$ namely
      $a = -29,$ $d = 2.$



      By the distributive law, we have equality of the sum of the first $12$ terms and the sum of the first $18$ terms whenever $a = -29k$ and $d = 2k$
      for some constant $k.$
      And then the sum of the first $21$ terms is $-189k.$



      So the sum of the first $21$ terms could be any real number if $k$ is allowed to be any real number. It can be any multiple of $189$ if the terms must all be integers.



      I suspect an error in the transcription of the question, possibly before it was presented to you.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        An arithmetic progression such that the sum of the first $12$ terms equals the sum of the first $18$ terms:
        begin{multline}
        (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
        (-17) + (-15) + (-13) + (-11) + (-9) + (-7) = -216,\
        end{multline}
        begin{multline}
        (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
        (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
        (-5) + (-3) + (-1) + 1 + 3 + 5 = -216. \
        end{multline}



        The sum of the first $21$ terms of the series is



        begin{multline}
        (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
        (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
        (-5) + (-3) + (-1) + 1 + 3 + 5 +7 + 9 + 11 = -189. \
        end{multline}



        So it is possible to do this with non-zero $a$ and $d,$ namely
        $a = -29,$ $d = 2.$



        By the distributive law, we have equality of the sum of the first $12$ terms and the sum of the first $18$ terms whenever $a = -29k$ and $d = 2k$
        for some constant $k.$
        And then the sum of the first $21$ terms is $-189k.$



        So the sum of the first $21$ terms could be any real number if $k$ is allowed to be any real number. It can be any multiple of $189$ if the terms must all be integers.



        I suspect an error in the transcription of the question, possibly before it was presented to you.






        share|cite|improve this answer









        $endgroup$



        An arithmetic progression such that the sum of the first $12$ terms equals the sum of the first $18$ terms:
        begin{multline}
        (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
        (-17) + (-15) + (-13) + (-11) + (-9) + (-7) = -216,\
        end{multline}
        begin{multline}
        (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
        (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
        (-5) + (-3) + (-1) + 1 + 3 + 5 = -216. \
        end{multline}



        The sum of the first $21$ terms of the series is



        begin{multline}
        (-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \
        (-17) + (-15) + (-13) + (-11) + (-9) + (-7) +{} \
        (-5) + (-3) + (-1) + 1 + 3 + 5 +7 + 9 + 11 = -189. \
        end{multline}



        So it is possible to do this with non-zero $a$ and $d,$ namely
        $a = -29,$ $d = 2.$



        By the distributive law, we have equality of the sum of the first $12$ terms and the sum of the first $18$ terms whenever $a = -29k$ and $d = 2k$
        for some constant $k.$
        And then the sum of the first $21$ terms is $-189k.$



        So the sum of the first $21$ terms could be any real number if $k$ is allowed to be any real number. It can be any multiple of $189$ if the terms must all be integers.



        I suspect an error in the transcription of the question, possibly before it was presented to you.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 21:14









        David KDavid K

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