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Given $3$ arithmetic progressions with same difference $d$:

$(1)$ $a_1,a_2,a_3,...$ $(2)$ $b_1,b_2,b_3,...$ $(3)$ $c_1,c_2,c_3,...$.

defined: $S_n=a_1+a_2+a_3+...+a_n$

$T_n=b_1+b_2+b_3+...+b_n$

$R_n=c_1+c_2+c_3+...+c_n$

Need to prove that if $a_1,b_1,c_1$ is an arithmetic progression so $S_n,T_n,R_n$ is also an arithmetic progression.

I tried to use the fact that $2b_1=a_1+c_1$ and $a_2-a_1=b_2-b_1=c_2-c_1=d$ but i'm stuck

Any idea?

Thanks.

EDIT - $a_n,b_n,c_n$ are arithmetic progressions

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  • $\begingroup$ $2T_n = S_n+R_n$. $\endgroup$ – njguliyev Sep 27 '13 at 16:52
  • $\begingroup$ Also, just to clarify, do you mean that the sequence $S_n$ is an arithmetic progression, or that the 3 terms $S_i, T_i, R_i$ form an arithmetic progressions for all $i$. There is slight ambiguity in your statement. $\endgroup$ – Calvin Lin Sep 28 '13 at 14:11
  • $\begingroup$ @CalvinLin if $a_1,a_2,a_3,...$ is a arithmetic progression so of course $S_n=a_1+a_2+a_3+...+a_n$ is arithmetic series isn't? because it's the sum of a finite arithmetic progression. we need to prove that the 3 terms form an arithmetic progression. $\endgroup$ – greeno Sep 28 '13 at 18:49
  • $\begingroup$ No, let me clarify. If $S_n = 1, 0, 1, 0, \ldots$, $T_n = 2, 1, 2, 1, \ldots$ and $R_n = 3, 2, 3, 2, \ldots$, then the three terms $S_i, T_i, R_i$ form an AP (either $1-2-3$ or $0-1-2$), but neither none of the sequences themselves are an AP. $\endgroup$ – Calvin Lin Sep 28 '13 at 19:30
  • $\begingroup$ $S_n$ is a sum. if $a_1,a_2,...=1,0,...$ which is a AP so $S_n=1+0+...$ which just the sum of it. otherwise i can't see what do you mean. $\endgroup$ – greeno Sep 28 '13 at 19:49
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Because $a_i$'s, $b_i$'s and $c_i$'s are arithmetic progression, we have $a_i=a_1+(i-1)d$, $b_i=b_1+(i-1)d$ and $c_i=c_1+(i-1)d$.

Therefore $a_i-b_i=a_1-b_1$ and $b_i-c_i=b_1-c_1$. By applying this result to the difference of $S_n,T_n$ and $R_n$, we obtain the following result: $$ T_n-S_n=n(b_1-a_1) \\ R_n-T_n=n(c_1-b_1) $$ But $a_1, b_1$ and $c_1$ form an arithmetic progression and therefore $a_1-b_1=b_1-c_1$. So we can see that $T_n-S_n=R_n-T_n$, which means that $S_n,T_n, R_n$ is an arithmetic progression.

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  • $\begingroup$ @CalvinLin, I get the problem like this. $a_1,b_1$ and $c_1$ make an arithmetic series and the question is to prove $S_n,T_n, R_n$ is also arithmetic. My suggestion was to show that $S_n-T_n=T_n-R_n$. Did I miss anything? $\endgroup$ – Arash Sep 27 '13 at 19:18
  • $\begingroup$ Ummm I see your point. I think OP means progression rather than series. Because he talks about having the same difference $d$ and also $a_2-a_1=d$. I think in the OP, the term "series" is better to be changed with "progression". $\endgroup$ – Arash Sep 27 '13 at 23:36
  • $\begingroup$ On the other hand, if OP really means Arithmetic series, I do not thing the result changes. $\endgroup$ – Arash Sep 27 '13 at 23:46
  • $\begingroup$ @CalvinLin - sorry it's a progression. ill edit. $\endgroup$ – greeno Sep 28 '13 at 10:51
  • $\begingroup$ can anyone help after my edit? $\endgroup$ – greeno Sep 29 '13 at 18:09

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