Question

What is the sum of a 38–term arithmetic sequence where the first term is 14 and the last term is 154?
a) 2660
b)2850
c)2926
d)3192

Answer 1

Well first you have to find out what the difference between terms is. Do you know how to do that?

Answer 2

O wait no you don't sorry didnt read your question carefully.

Answer 3

You don't have to.

Answer 4

do you use the arithmetic sequence formula=a1+(n-1)d substitute the values and find d??

Answer 5

Well, I'd rather have you do it the following way.
Let's focus on how to get the sum. We have an arithmetic sequence, with a term \[t_n\] being the nth term. Thus the thing you're looking for is:
\[t_1+t_2+...t_34\]

Answer 6

Correction:
\[t_{34}\]

Answer 7

Good so far?

Answer 8

Yeah

Answer 9

Now I propose we do this: Pair the first and last terms, then second and second to last, etc, etc. This is the commutative and associative property of addition, so it's got to be valid:
\[t_1+t_{34}+t_2+t_{33}...=\]
\[(t_1+t_{34})+(t_2+t_{33})...\]

Answer 10

Good?

Answer 11

Um..yes

Answer 12

Now let's say you have the mean m such that
\[m = \frac{t_1+t_{34}}{2}\]

Answer 13

Okay. Now let's say I look at
\[\frac{t_2+t_{33}}{2}\]

Answer 14

Use the formula you mentioned earlier with difference d between terms to substitute t2 and t33 with t1 and t34

Answer 15

Well?

Answer 16

so the its 154+14/2?

Answer 17

No, no. Let's go symbolic. Replace t2 and t33 using only t1, t34, and d.

Answer 18

huh?ok sorry but now im really lost..

Answer 19

Okay. Do you agree that
\[t_2 = t_1-d\]and\[t_33=t34-d\], right?

Answer 20

yes..

Answer 21

wait no made a couple typos:
\[t_2 = t_1+d\]and\[t_{33}=t_{34}-d\]

Answer 22

There. Now, going back to our little example of
\[\frac{t_2+t_{33}}{2}\]

Answer 23

Replace t2 and t33...

Answer 24

Well, what do you get?

Answer 25

2660?

Answer 26

Look, I'm trying to lead you to the answer. Stay symbolic. K, I'll just show you. So we had:
\[m=\frac{t_1+t_{34}}{2}\]What I wanted you to do was:
\[\frac{t_2+t_{33}}{2}=\frac{t_1+d+t_{34}-d}{2} = \frac{t_1+t_{34}}{2}\]agree?

Answer 27

yes

Answer 28

Then, you should see that
\[\frac{t_3+t_{32}}{2}\] also equals m

Answer 29

By using
\[t_3 = t_2 + d\]and\[t_{32}=t_{33}-d\]and then you get\[\frac{t_{33}+d-d+t_2}{2}=\frac{t_{33}+t_2}{2} =\frac{t_{34}+t_1}{2} = m\]

Answer 30

Agree?

Answer 31

yes

Answer 32

Okay. Then, going back to my sequence...
\[(t_1+t_{34})+(t_2+t_{33})...\]

Answer 33

Which is still the original sum you're looking for, I do the following, multiplying by 1.
\[(t_1+t_{34})+(t_2+t_{33})...=\]\[1*((t_1+t_{34})+(t_2+t_{33})...)=\]\[(2/2)*((t_1+t_{34})+(t_2+t_{33})...)=\]\[2*((t_1+t_{34})/2+(t_2+t_{33})/2...)=\]\[2*(m+m...)=\]

Answer 34

Do you know how many m's there will be summed together?

Answer 35

look the question has 4 possible answers a)2660 b)2850 c)2926 and d)3192

Answer 36

Well, I could simply answer for you but this proof is much more important; that is, only if you want to learn...

Answer 37

Okay. Anyway. Notice that there were 34 numbers. I PAIRED them. How many PAIRS are there - each pair turned into an "m"

Answer 38

17?

Answer 39

Yes! 17! Bettern known as 34/2!
\[2*(m+m+m...)=2*(m\frac{34}{2})=34 m\]

Answer 40

Remember, this was originally
\[t_1+t_2...+t_{34}\]

Answer 41

So there's your answer. Take the mean of t1 and t34, and then multiply by the number of terms, 34!

Answer 42

ok..

Answer 43

Well?

Answer 44

but how do u get the mean of t1 and t34?

Answer 45

I'm really confused -.-

Answer 46

t1 is the first term, and t34 is the last term. How's it that hard?

Answer 47

In fact, for any arithmetic sequence of length n, the sum is always
\[n*mean = n\frac{t_1+t_n}{2}\]

Answer 48

I still dont get it!!! plzz help!

Answer 49

Okay. Let's try a different approach... Imagine you have a 4-term arithmetic series:
\[t_1+t_2+t_3+t_4\]

Answer 50

You know that between each term there's a difference d.

Answer 51

Forget it..I got it thnxz anyways

Answer 52

The answer is 3192

Answer 53

You use the formula Sn=n/2(a1+an)

Answer 54

Thats what I proved!!!

Answer 55

Where n is 38, a1 is 14 and an is 154

Answer 56

What I said earlier, verbatim:
In fact, for any arithmetic sequence of length n, the sum is always
\[n∗mean=n\frac{t_1+t_2}{2}\]

Answer 57

t_n not t_2**

Answer 58

ok....

Answer 59

Your equation for Sn is the exact same, but uses a instead of t... But fine. If you insist on memorizing a formula, then you're ok, right?

Answer 60

Yah I'm ok i hav a pictographic memory I can't help but memorize

Answer 61

Pretty sure that's not proven to exist but w/e.

Answer 62

Whatever...

Answer 63

K anyway remember to close the question!