Alright, these are messing me up, so...

4. The sum to infinity of a geometric series if four times its first term. Find the common ratio.

3. How many terms of the arithmetic series 15+13+11... are required to make a total of -36?
3) a = 15 d = -2

Sn = n/2 (a + (n - 1) d)
-36 = n/2 (15 + (n - 1) -2)
-72 = n (15 - 2n + 2)
2n^2 - 17n + 72 = 0
(2n - 9)(n - 8) = 0

n = 8 or n = 9/2
FUCK YOU! GET PUMPED!
Quote by Craigo
Alright, these are messing me up, so...

4. The sum to infinity of a geometric series if four times its first term. Find the common ratio.

3. How many terms of the arithmetic series 15+13+11... are required to make a total of -36?

4.

S-inf = a/1-r

a/1-r=4a
a = 4a-4ar
1=4-4r
r=3

Edit:
Damn;
4r=3
r=3/4
Last edited by bequickorbedead at May 30, 2008,
4.

S = u/(1-r)

in this case becomes
4u = u/(1-r)
4u(1-r) = u
=> 1-r = 1/4
r = 3/4

4.

S-inf = a/1-r

a/1-r=4a
a = 4a-4ar
1=4-4r
r=3

Dude, condition for Sum to infinity to hold is that r < 1.

EDIT again, sorry, seems you realised yourself (-:
Last edited by Mad_BOB at May 30, 2008,
haha brain hurts from reading the questions

i took the exam 2 weeks ago pheeew

EDIT: core 2?
Quote by SuperBlob
3) a = 15 d = -2

Sn = n/2 (a + (n - 1) d)
-36 = n/2 (15 + (n - 1) -2)
-72 = n (15 - 2n + 2)
2n^2 - 17n + 72 = 0
(2n - 9)(n - 8) = 0

n = 8 or n = 9/2

i would like to correct this one

3) a = 15 d = -2

Sn = n/2 (2a + (n - 1) d)
-36 = n/2 (30 + (n - 1) -2)
-72 = n (30 - 2n + 2)
-72 = 32n -2n^2
2n^2 - 32n - 72 = 0
(2n + 4)(n - 18) = 0

n = 18 or (n = -2 which it cannot be)
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Ah crap.

I knew it was off with it giving 2 answers
FUCK YOU! GET PUMPED!