Question

The sum Sn of finite arithmetic series of n terms is Sn= n/2(A1+An) where A1 is the first term and An is the nth term.
a) show that Sn=na1+ n(n-1)/2 d by replacing An with its value in term of A1, n, and d in the above formula
b) explain why Sn=na1+ n(n-1) d makes sense by explaining how you can extract each of na1 and n(n-1)/2 d from the sum A1+(A1+d)+(A1+2d)+...+(a1+(n-1)d)

Answer 1

@rebeccaxhawaii , practice for u bb

Answer 2

can you take a screenshot of the question.

Answer 3

someone please help i have no idea how to do this

Answer 4

Hint:
\[\Large S_n = \frac{n}{2}\left(a_1 + a_n\right)\]
\[\Large S_n = \frac{n}{2}\left(a_1 + {\color{red}{a_n}}\right)\]
\[\Large S_n = \frac{n}{2}\left(a_1 + {\color{red}{a_1+d(n-1)}}\right)\]
I'm using the nth term formula (for arithmetic sequences) which is
\[\Large a_n = a_1+d(n-1)\]

Answer 5

it still makes no sense at all to me

Answer 6

Do you see how I replaced the \(\Large a_n\) with \(\Large a_1+d(n-1)\) ?

Answer 7

The first question is trying to get us to show that Sn=n(a1+an)/2=na1+n(n-1)d/2.

Answer 8

do you agree?

Answer 9

yes i do and yes i agree that it is showing that

Answer 10

If you expand the sum Sn - the sum up to the nth term - you would get some thing that looks like this:

Answer 11

Finally, you just need to substitute an into Sn=n(a1+an)/2 and it should produce Sn= na1+n(n-1)d/2 if you expand the terms.