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I've got a question regarding a mathematical problem I've been trying to solve.

How can I find the first term and the common rule of this arithmetic sequence with the following info:

The common difference ($d$) = $11$

The sum of the first 40 terms of the sequence = $8700$

So, how can I find the first term and the common rule of this sequence.

I hope I can get some help because I've been struggling trying to solve this problem for quite a while.

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2 Answers 2

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Hint: If the starting term were $0$, the sum of $40$ terms would be eleven times the sum of the numbers from $0$ to $39$. Can you do that? Then increasing the starting term by $1$ increases all the terms by $1$ and therefore the sum by $40$.

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  • $\begingroup$ Still couldn't quite get the correct answer. Is there any way to make an equation and solve it that way? $\endgroup$ Commented Mar 27, 2017 at 20:46
  • $\begingroup$ Yes there is. Your series is $a, a+1, a+2 \ldots, a+39$ Adding them up gives $40a+\sum_{i=0}^{39}i=8700$ That is why I asked if you could do the last sum. $\endgroup$ Commented Mar 27, 2017 at 20:51
  • $\begingroup$ Now that you have changed the difference, your series is $a, a+11, a+22, \ldots a+429$. Adding them up gives $40a+11\sum_{i=0}^{39}=8700$ $\endgroup$ Commented Mar 28, 2017 at 1:20
  • $\begingroup$ Ah, got it now. Thanks a lot for the help. $\endgroup$ Commented Mar 28, 2017 at 4:09
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Since common difference is $11$, we let the first term be $x$. Then the sequence is $x+(x+11)+(x+2\times 11)+...+(x+39\times 11)=8700$. Then, the total number of $11$ is $1+2+3+...+39=\frac{39\times 40}{2}=780$. The sum of these $11$ is then $11\times 780=8580$. Now we have $40x=8700-8580=120$. We have $x=3$. Note that this solution probably isn't very relevant since the question was so long ago, but please accept of it helps.

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