# Thread: Arithmetic series: sum of 7 + 14 + 21 + 28 + ... + 98

1. ## Arithmetic series: sum of 7 + 14 + 21 + 28 + ... + 98

Hi, I have no idea how to set these problems up:

Find the sum of the arithmetic series.

1) 7 + 14 + 21 + 28 + ... + 98

2) What is the sum of the mulitples of 3 between 3 and 999, inclusive?

2. Well as you can see the sequence is formed by your 7 multiplication tables:

7 * 1 = 7

7 * 2 = 14

....

So the last number in the sequence is 98 which is 14 * 7.

So there's 14 numbers in the sequence.

Starting from the first number and the very last number the sum is 105.

Then to the second number and 13th number the sum is 105.

Notice the patern.

There is 2 numbers in the sequence that makes 105 for each corresponding pair. You have 7 pairs.

3. ## Re: Arithmetic series

Originally Posted by Laura54321
What is the sum of the mulitples of 3 between 3 and 999, inclusive?
It is important to point out a matter of wording in this question.
The word between means exactly that. In this case it means 6 to 996.
That is a commonly agreed upon editing standard; I have served on many editing committees for mathematics contest questions. If one means to ask the sum of all multiples of three from 3 to 999 then note the wording I used: from…to…

Adding the word “inclusive” unfortunately does not meet the objection.
$\L \sum\limits_{k = 1}^{333} {3k} = 3\sum\limits_{k = 1}^{333} k = 3\left( {\frac{{\left( {333} \right)\left( {334} \right)}}{2}} \right)$.

4. ## Re: Arithmetic series: sum of 7 + 14 + 21 + 28 + ... + 98

Originally Posted by Laura54321
Hi, I have no idea how to set these problems up:
WHY do you have no idea?
You were absent from math classes for reasons beyond your control?
You were unable to google "arithmetic series" due to unforeseen circumstances?

5. ## Re: Arithmetic series: sum of 7 + 14 + 21 + 28 + ... + 98

[quote="Laura54321"]Hi, I have no idea how to set these problems up:

Find the sum of the arithmetic series.

1) 7 + 14 + 21 + 28 + ... + 98

2) What is the sum of the mulitples of 3 between 3 and 999, inclusive?

The sum of an arithmetic series is given by S = n(a + L)/2 where a == the first term, L = the last term and n = the number of terms.

For problem 1),, a = 7, L == 98 and n = 14.
Thus, S = 14(7 + 98)/2

For problem 2), a = 3, L = 999 and n = 333.
Thus, S = 333(3 + 999)2.

In case your statement "between 3 and 999" actually means from 6 to 996, S = 331(6 + 996)/2.

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