Question

Give the sum of this arithmetic series:
2+4+6+...+100

Answer 1

first, find out how many terms there are by doing
\[a_n = a_ 1 + (n-1)d\]
an is the last term in this case 100
a1 is the first term, in this case 2
d is the common difference
and n is what you're looking for

Answer 2

can you try doing it?

Answer 3

solve for n

Answer 4

sum = (n/2) [ a + l]

Answer 5

\[{n \over 2}\left(a_1 + a_n \right)\]First, find the number of terms.\[a_n = a_1 + (n - 1)d\]\[a_n = 2 + 2(n - 1)\]solve for n

Answer 6

161 terms, right?

Answer 7

lol no

Answer 8

Oh I see what i did, give me one sec man

Answer 9

sure

Answer 10

take yur time

Answer 11

number of terms n = 50
a = 2 and l last term = 100

Answer 12

Can you show your work on how you got the 50 for n?

Answer 13

if you substitute the values
\[\implies 100 = 2 + (n-1)2\]
you get that right?

Answer 14

Right the common difference is 2

Answer 15

right. so if you distribute 2 into n-1 you get
\[\implies 100 = 2 + 2n - 2\]
right?

Answer 16

so 100=2n
n=50

Answer 17

right

Answer 18

Then we need to use the sum of an aritmetic series formula, correct?

Answer 19

so now, you can use the formula \[S_n = \frac{n(a_1 + a_n)}{2}\]

Answer 20

yes you're right

Answer 21

Thank you. Hold on real quick as I complete the problem

Answer 22

sure

Answer 23

So you would set it up as this, correct: s(sub 50)=50(2+100)/2

Answer 24

yes!

Answer 25

And get (25)(102), and get 2550?

Answer 26

right

Answer 27

Can also think of it like this:
2+4+6+8+...+100 = (2+100)+(4+98)+(6+96)+(8+94)+...(50+52).
If you can't remember the formula, this is how you derive it.

Answer 28

Interesting, thanks guys
Hold one, I have one more to try

Answer 29

sure

Answer 30

Ok, the next problem is 1+5+9+...+401
So, first thing I did was use the arithmetic sequence formula a to the nth term=a1+(n-1)d So i plugged in 401=1+(4n-4) Ended up getting 101 for n
Then I used the sum of an aritmetic series formula s(101)=101(1+401)/2
Ended up with 40602/2 And got 20301 as the sum of all 101 terms. Is that correct Igbasallote?