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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?

mac07
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  • @Nick Using the formula ,Sum of n terms = n(2a + (n-1)d)/2 – mac07 Aug 13 '16 at 05:36
  • 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 $\sum_{i=1}^{21} x_i = \sum_{i=1}^{21} 0 = 0$.

     – benguin Aug 13 '16 at 05:43
  • 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 – Nick Aug 13 '16 at 05:44
  • @Nick How did you get S14+3a+57d=21a+210d=S21 ? – mac07 Aug 13 '16 at 05:57
  • $S_{21}=S_{18}+a_{19}+a_{20}+a_{21}=S_{14}+a_{19}+a_{20}+a_{21}$ – Nick Aug 13 '16 at 06:01
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    Hint: get the second equation by equating the sum of $x_{13} to x_(18}$ to zero. – Moti Aug 13 '16 at 06:04
  • @Moti,@Nick sum of 13th term to 18th term is zero. So S21 should be equal to S12 + a19 + a20 + a21 = S18 + a19 + a20 + a21 right ? But how did you get S21 = S14+a19+a20+a21 ? – mac07 Aug 13 '16 at 06:18
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    @benguin I finally got two equations : 6a + 87d = 0 and 2a + 27d = 0. Solve these equations give a=d=0 – mac07 Aug 13 '16 at 07:16

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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.

David K
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