Question

Find the indicated nth partial sum of the arithmetic sequence. 1.9, 4.8, 7.7, 10.6, ...., n=20.
Show your work.

Answer 1

you will need to determine the common difference between terms then

Answer 2

once you determine the common difference "d", there is a general formula for a partial sum\[S_n=\frac{n(a_1+a_n)}{2}\]
\[S_n=\frac{n(a_1+(a_1+d(n-1))}{2}\]
\[S_n=\frac{n(2a_1+dn-d)}{2}\]
since you know the first term a1, and the nth or 20th element

all you really needs is the common difference "d"

Answer 3

so.. \[S _{n}=551\]
? @amistre64

Answer 4

maybe ...

4.8
1.9
---
2.9 = d

a20 = 1.9+2.9(19) = 57

20(1.9+57)
---------- = 10(58.9) = 589
2

Answer 5

woopss. somehow i subtracted wrong finding d. ha. okay, i get it now. thanks.

Answer 6

@amistre64 can you help me with this one?

Answer 7

1n - 3n

Answer 8

\[\sum_{n=1}^{299}n-3\sum_{n=1}^{k} n\]
299/3 = 99 + .... so k=99
\[\sum_{n=1}^{299}n-3\sum_{n=1}^{99} n\]

Answer 9

essentially

\[\sum_{n=1}^{299}n-3\sum_{n=1}^{99} n\]
\[\sum_{n=1}^{99}n-3\sum_{n=1}^{99} n+\sum_{n=100}^{299}n\]
\[-2\sum_{n=1}^{99}n+\sum_{n=100}^{299}n\]
\[\frac{99(-2-198)}{2}+\frac{200(100+299)}{2}\]

Answer 10

or instead of trying to get fancy :)\[\frac{299(1+299)}{2}-3\frac{99(1+99)}{2}\]
\[299(150)-3(99)(50)\]

Answer 11

so the sum of the integers that are not multiples of 3 is 30,006?

Answer 12

14850

Answer 13

I'm lost.

Answer 14

3(99) = 297 so all the multiples of 3 from 1 to 99 are subtracted

3( 1 + 2 + 3 + 4+...+99)
+99+98+97+96+...+ 1
-------------------------
3(100+100+100+100+...+100)
|------- 99 times----------| which is 2 times more than we want

99(100) divided in half is 99(50)

Answer 15

we are basically wanting to count the set twice in a convenient manner

take the sum of the digits from 1 to 299

1 + 2 + 3 + 4 +...+296+297+298+299 , lets add this to
299+298+297+296+...+ 4 + 3 + 2 + 1 itself backwards
-----------------------------------------
300+300+300+300+...+300+300+300+300

thats 300, 299 times. we only need half of that count

Answer 16

Yeah.. im still completely lost :/

Answer 17

do you agree that if we add up all the integers from 1 to 299, twice,
we get 300 * 299 ?

Answer 18

look at the setup i did, and see if you can convince yourself that its true :)

Answer 19

we can try it on smaller sets if need be

1+2+3
3+2+1
------
4+4+4 = 3*4 = 12, which is twice what we need
12/2 = 6

1+2+3 = 6

Answer 20

2 + 4+ 6+ 8+10
10+ 8+ 6+ 4+ 2
-----------------
12+12+12+12+12 = 5*12 = 60, which is twice what we need
60/2 = 30

2 + 4 + 6 + 8 + 10 = 30

Answer 21

i think i know where im getting lost.
after you say
1+2+3
3+2+1
-------
then you say 4+4+4 but where did you get this?

Answer 22

add the columns ...

Answer 23

3+1 = 4
2+2 = 4
3+1 = 4

Answer 24

each column amounts to the sum of the first and last numbers of the arithmetic sequence

Answer 25

oh. okay. i see where you're coming at from that now.

Answer 26

since the each column is the equal to the sum of the first and last values; we only really need to know the first and last values ... times the number of "numbers" there are.

3 * (1+3) = 3*4 = 12 ... 12/2 = 6

5*(2+10) = 5*12 = 60 .... 60/2 = 30

299(1+299) = 299*300 = 89700 ..... 89700/2 = 44850

Answer 27

subtract from that the multiples of 3, from 1 to 99

Answer 28

3 * (1+99) * 99 = 3 * 100 * 99 ..... and divide by 2: 14850

44850
-14850
---------
30000