Arithmetic Series - Easy questions

Hai,

I've totally forgtten how to do these if anyone would care to explain :smile:

9c) The rth term of an arithmetic series is (2r-5)

Show that E (weird symbole with "n" on top of the "E" and r=1 under it)

E (2r - 5) = n(n-4)

It's the C1 book page 144 question 9c :smile:

Also, question 11b)

A sequence of a1, a2, a3 is defined by a1 = 3

3a + 5 was the formula and i figured out

a1 = 3 a2 = 4 a3 = 7

Caculate the value of E (weird symbole with "5" on top and r=1 under it)

- Thanks

Reply 1

EEngWillow

Original post by Sadbunny11

Hai,

I've totally forgtten how to do these if anyone would care to explain :smile:

9c) The rth term of an arithmetic series is (2r-5)

Show that E (weird symbole with "n" on top of the "E" and r=1 under it)

E (2r - 5) = n(n-4)

It's the C1 book page 144 question 9c :smile:

Your E (weird symbol with "n" on top and r = 1 under) is a sigma. It denotes the sum of a set of terms in the series from r = 1 to n.

1) The sum of n terms in an arithmetic series is given by

\frac{n}{2}(2r_1 + (n-1)d)

r_1

being the first term, and d the common difference.

2) Again you're being asked to calculate the sum of the first n terms of a series. I'd point out here that your a1, a2 and a3 don't have a common difference, so I'd look at recalculating those.

Reply 2

Potassium^2

The E is called sigma

Reply 3

Sadbunny11

Hai - thanks for the reply.

I just noticed that for 11b) the one that hasn't got a common difference it hasn't actually stated its a arithmetic series (i just assumed it D: - my mistake).

Is the only way to add each step up then? Since i dont believe i was given a formula.

Also, thanks for clearing up 9c) aswell, but how would i go from...

n/2 [2r + (n-1) d]

to

(2r-5) = n(n-4)

A tad confused on that :s-smilie:

- Thanks

Reply 4

EEngWillow

Original post by Sadbunny11

Also, thanks for clearing up 9c) aswell, but how would i go from...

n/2 [2r + (n-1) d]

to

(2r-5) = n(n-4)

A tad confused on that :s-smilie:

- Thanks

You aren't trying to equate (2r-5) to n(n-4). You're saying that the sum of the first N terms in the series (2r-5) is equal to n(n-4)

You have the equation for the sum of terms in any arithmetic sequence (n/2 [2r + (n-1) d]).

Basically,

Sigma (2r-5) = n/2 [2r + (n-1) d]
Get your common difference d, and your first term r, and you're sorted with a bit of simplifying.

EDIT:

For the other question, what is actually given to you in the question? The formula?