My first item = 100. Each following item is = +100 (so my 5th item is worth 500)
How do I find the sum total of items 21 to 57?
...You know other than having a large sheet with all the items and pulling = sum ( range)
$\begingroup$
$\endgroup$
-
1$\begingroup$ Do you know the formula for the sum of an arithmetic series? It would be the easiest way to do this $\endgroup$ – Osama Ghani Jun 22 '17 at 8:35
-
3$\begingroup$ No, but now that you've put a name on what I'm looking for, I'll see how far I can self educate on the internetz. $\endgroup$ – helena4 Jun 22 '17 at 8:38
$\begingroup$
$\endgroup$
The average of the items between 2100 and 5700 is 3900. If we multiply the average of the items by the number of items we should get the sum of all the items. There are 37 total items between 21 and 57 so we multiply $$37*3900$$ which equals 144,300
-
6$\begingroup$ This is a good technique, you should point out though how to calculate the average. By definition it would be $(2100+2200+\dots +5700)/37$ but the sum in the nominator is what you want to calculate. However the average is also $(2100+5700)/2$ since we can pair the items together to give the same sum: $2300+5500=2200+5600=2100+5700$. $\endgroup$ – Mathematician 42 Jun 22 '17 at 8:50
-
-
$\begingroup$
$\endgroup$
That's an arithmetic progression. You are given a progression $a_1=100$, $a_2=200$, ...
The sum of the first $n$ terms is given by: $$S_n=\frac{n}{2}(2a_1+(n-1)d),$$ $d$ being the step size ($100$ in your case). Since you want to know the sum of items from 21 to 57, you just need to compute $S_{57} - S_{20}$.
$\begingroup$
$\endgroup$
You can use the formula for the sum of an arithmetic series, which is: $$(n/2)(a_1 + a_n),$$
where $a_1$ is the first term, $a_n$ is the last term and $n$ is the number of terms.
We know that the first term is $21*100$, and by the counting principle there are $(a_n - a_1)+1$ terms, so plugging in the numbers we get $(57-21) + 1$ terms or $37$ terms.
Plugging the numbers again gives $[37/2][21*100+57*100]$, which you can compute from there.
$\begingroup$
$\endgroup$
You can write this as $$\sum_{i=21}^{57}100i=100\sum_{i=21}^{57}i.$$
Now, it's well known that $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$. Observe that $$100\sum_{i=21}^{57}i=100(\sum_{i=1}^{57}i-\sum_{i=1}^{20}i)=100(\frac{57\cdot 58}{2}-\frac{20\cdot 21}{2})=50(57\cdot 58-20\cdot 21).$$
Some more calculations yield $$50(57\cdot 58-20\cdot 21)=100(57\cdot 29-10\cdot 21)=300(19\cdot 29-10\cdot 7)=300(481)=144300$$
answered Jun 22 '17 at 8:36
$\begingroup$
$\endgroup$
An old trick is to write $$S=\underbrace{2100 + 2200 + 2300 + \cdots + 5600 + 5700}_{57-21+1=37 \textrm{ terms}}$$ Summing backwards $$S = \underbrace{5700 + 5600 + 5500 + \cdots + 2200 + 2100}_{57-21+1=37 \textrm{ terms}}$$ Adding vertically $$S+S = \underbrace{7800 + 7800 + 7800 + \cdots + 7800 + 7800}_{57-21+1=37 \textrm{ terms}}$$ $$2S =37\times 7800 =288,600$$ $$S = \boxed{144,300}$$
$\begingroup$
$\endgroup$
What you're describing is an Arithmetic Sequence. So we can use the formula:
Σn = n/2(2a+(n-1)d),
where Σ = "the sum of", n = the last term, a = the first term, and d = the difference
to find the sum of the arithmetic sequence up to the term n. However as we are trying to find the sum of terms between 21 and 57, a simple way to do this is to call the 21st term as the 1st term a, and the 57th term would be 36 terms above the term a (57-21, although we include the term a here so the n value we will use is actually n = 37)
Sub in values: a = 21(100)=2100, d = 100, n = 37.
So ∑n = 37/2(4200+(37-1)100)
Therefore ∑ = 144300
Hope this helps!
-
$\begingroup$ Toby Mak's solution is a lot simpler than mine, but hopefully this other method provides some deeper understanding :) $\endgroup$ – Ivan Galt Jun 22 '17 at 9:11
-
$\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jun 22 '17 at 10:13
