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I realize there are similar stacks to my question such as: How many summands are there

Although I require further clarifications to understand.

Consider the sequence: 4 + 11 + 18 + 25 + ... + 249.

1) How many summands are in the sum.

2) Compute the sum.

aitía
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  • https://en.wikipedia.org/wiki/Arithmetic_progression – haqnatural May 27 '17 at 21:22
  • Do you know the general term of the summands? In general do you know what kind of sequence is this (well what you wrote is a sum,but terms from that sum form a sequence)? Also where did you find this problem, if it was from textbook you could consider looking at the definitions of that chapter. – kingW3 May 27 '17 at 21:23
  • By general term I believe you mean an = a1 + (n - 1)d. Yes it is an arithmetic sequence with a common difference of 7. You are correct, I am referencing within a chapter. – aitía May 27 '17 at 21:29
  • Wolfram alpha calculate the sum directly and additionally displays the sigma sign notation. http://www.wolframalpha.com/input/?i=4+%2B+11+%2B+18+%2B+25+%2B+...+%2B+249 – callculus42 May 27 '17 at 21:38
  • @aitía Yes, I assume you know that $a_1=4$ and $d=7$ now consider that $a_n=4+(n-1)\cdot 7=249$ so we get $245=(n-1)\cdot 7$ which gives $n-1=\frac{245}{7}=35$ so $n=36$ (36 terms),from this it should be easy to calculate the sum with the formula $S_n=\frac{n}{2}(a_1+a_n)$. – kingW3 May 27 '17 at 22:05
  • I wish to say thanks, all of you for these explanations, I now understand fully! – aitía May 29 '17 at 01:44

2 Answers2

1

By simple investigation, you can observe that the sequence you're dealing with is an arithmetic progression, hence if you name it $a_n$, then you're interested in summing the sequence $a_n$ defined by:

$$ a_n=4+7n, n\in\mathbb N $$

And to know how many summands there are, yoi solve for $n$ such that:

$$ a_n=249=4+7n $$

Hence $n=35$, so there are $36$ summands.

The sum is simply:

$$ a_0+a_1+...a_{35}=\sum_{n=0}^{35}4+7n=4\times 36+7\times\frac{35\times 36}{2}=4554 $$

Oussama Boussif
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Since there are $36$ terms as observed by the other answers the sum of the arithmetic progression is $$ \frac{\text{number of terms}\left(\text{first}+\text{last}\right)}{2}=\frac{36(4+249)}{2}=18(253)=4554. $$

Sri-Amirthan Theivendran
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