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If we have a geometric sequence $a= (a_n)_n$ and an arithmetic sequence $b=(b_m)_m$. We can find the $n$th term of $a+b$ easily.

Now, suppose we have a sum of geometric sequnce and arithmetic sequnce (the ith term is added to the ith term). Can we do the process backwards and find the geometric sequnce and the arithmetic one? Here, we assume that we are given the sum as a sequence of numbers like $2 , 4 , 7 , 12 , 21 ...$( The sum of $1,2,3,4,5,...$ and $1,2,4,8,16,...$) but not given the $n$th term. How to do get the original sequences back in general?

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  • $\begingroup$ Let first term of AP be $a$, common difference $d$. Let $b$ be the first term of the GP, and $r$ the common ratio. Then if we know the first $4$ terms of the sum sequence, we get $4$ equations in $4$ unknowns. $\endgroup$
    – André Nicolas
    Aug 28, 2014 at 4:25
  • $\begingroup$ @AndréNicolas, I thought of that. but if we are given $20$ terms. then we got 20 equations! Is there any easier method? what if the sum is a sum of more than one geometric sequence? $\endgroup$
    – Love Coding
    Aug 28, 2014 at 4:29
  • $\begingroup$ If we are given $20$ terms, we throw the last $16$ away, they are not needed. A similar strategy should work for other "mixed" sequences. $\endgroup$
    – André Nicolas
    Aug 28, 2014 at 4:31

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We show how to find the two "mother" sequences, without paying attention to possible divisions by $0$ where things might break down.

Let the AP have first term $a$, common difference $d$, and let the GP have first term $b$, common ratio $r$. These are of course unknown. Let the first four terms of the "mixed" sequence be the known numbers $A,B,C,D$. These are your $c_1,c_2,c_3,c_4$.

Then we have

$$a+b=A;$$ $$a+d+br=B;$$ $$a+2d+br^2=C;$$ $$a+3d+br^3=D.$$ Use the first equation to eliminate $a$. We get $$d+br-b=B-A:$$ $$2d+br^2-b=C-A:$$ $$3d+br^3-b=D-A.$$ Use the first equation and the second, and the first and the third, to eliminate $d$. We get the two equations $$br^2-2br+b=C+A-2B;$$ $$br^3-3br+2b=D+2A-3B.$$ Take these last two equations and divide to eliminate $b$. We get $$\frac{r^3-3r+2}{r^2-2r+1}=K,$$ where $K$ is known.

Note that $\frac{r^3-3r+2}{(r-1)^2}=r+2$. So $r$ is now known and we can work backwards finding $b$, and then $d$, and then $a$.

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  • $\begingroup$ But it's not true that $\frac{r^3-3r+2}{(r-1)^2} = r+2$. use long division, you will find that there is a fraction. Also, just sketch$\frac{r^3-3r+2}{(r-1)^2}$ as if it were a function. the two expressions start being close to each other when $r$ is greater than $4$. but when it's less that $4$. the difference between $r+2$ and $\frac{r^3-3r+2}{(r-1)^2}$ is not small. Could you clarify that please? $\endgroup$
    – Love Coding
    Aug 28, 2014 at 8:24
  • $\begingroup$ There could be a mistake in the arithmetic earlier, I can look into it tomorrow if you find one. It would not make much difference to the structure of the calculation. However, $(r-1)^2$, that is, $r^2-2r+1$ does seem to divide $r^3-3r+2$. I have just done the long division, and even though it is very late, and my blood caffeine level is low, I still think the division yields $r+2$. If you like, plug in values. Take $r=2$. Then the top is $4$, the bottom is $1$, ratio $4$, which is $r+2$. Take $r=3$. The top is $20$, the bottom $4$, ratio $5$ which is $r+2$. And so on. $\endgroup$
    – André Nicolas
    Aug 28, 2014 at 9:17
  • $\begingroup$ You're right. they are equal! I made a mistake in the long division. I apologise for not being careful. Thank you very much :) $\endgroup$
    – Love Coding
    Aug 28, 2014 at 9:29
  • $\begingroup$ You are welcome. It is a good thing to check! $\endgroup$
    – André Nicolas
    Aug 28, 2014 at 9:32

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