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?
Best Answer
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.