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?
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?
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.
Try your best in everything and don't be scared to do something different.
It is important to point out a matter of wording in this question.Originally Posted by Laura54321
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.
But here is your answer:
[tex]\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)[/tex].
“A professor is someone who talks in someone else’s sleep”
W.H. Auden
WHY do you have no idea?Originally Posted by Laura54321
You were absent from math classes for reasons beyond your control?
You were unable to google "arithmetic series" due to unforeseen circumstances?
I'm just an imagination of your figment !
[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.
TchrWill
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