Sequences - geometric and arithmetic

[parksie]

Still don't get these, and have my Pure 2 retake next week.

Does anyone know how the equations for these work? I didn't understand most of the references I looked at.

[Guv]

Arithmetic progressions are fairly straight forward. If you forget the formula, you can always work out the answer by making pairs of terms. The simplest arithmetic progression is the first N integers. Suppose you were asked to add up all the integers from one to one thousand. Imagine them paired as follows. 1 1000 2 999 3 998 4 997 . . .It looks like 500 pairs, each of which adds up to 1001. Hence the sum of the series is 500*1001 = 500500.

The text book formula is N * (A + L)/2, where A is first term, L is last term, and N is the number of terms. Applied to the above you get 1000 * (1 + 1000) / 2.

Suppose you had the series: 8, 11, 14, 17, 20, 23, 26.

The formula method is 7 * (8 + 26) / 2 = 119. N = 7 terms, A = 8 is first, L = 26 is last.

Pairing is tuff because there is an odd number of terms, so pair the first 6 8 23 11 20 14 173 pairs, each adding to 31 or 3 * 31 = 93. Then add the last term: 93 + 26 = 119.

There is a formula which uses the difference between terms, and does not use the last term. I try to avoid this formula because I forget it, and it was messier than the above. I always seemed to be able to figure out what the last term was.

Does the above help?

Will see what I can think up about geometric series.

[parksie]

Thanks Guv, this really helps :)

Will try making a program to do it - I always understand things better when I write a program :confused:

[marnitzg]

I remember these? (Shame on you who don't!)

To find a specific term
Tn = A + (n-1)d

n is the term you're looking for. A is the first term and d is the difference between terms.
Just a simple Eg. 1;2;3;4;x
You want x which is the 5th term. First find the differene. So that would be T2-T1 which is 2-1=1 (Dah!)
So now your formula is:
T5 = 1 + (5-1)1
T5 = 5

Now the sum formula
from eg above
Sn = (n/2)(2a + (n-1)d)
S5 = (5/2)(2(1) + (4)1)
S5 = 15

Sn = (n/2)(a + L) has already been explained

Geometric
To find a specific term:
Tn = ar^(n-1)
a is the first term and r is the difference.
To find r: T2/T1

To find the sum of terms:
Sn = (a(r^n - 1))/(r-1) NB: if r > 1
Sn = (a(1 - r^n))/(1-r) if r < 1
OR
Sn = (a-r.L)/(1-r)
L is the last term

And finaly to find the sum to infinity:
S = a/(1-r)

[parksie]

Shame on me then :)

These are really helpful - considering I never got them first time round...that's why I'm retaking my Pure 2 exam :(.

[parksie]

Thanks a bundle, this was really helpful (just beasted the exam :))

[marnitzg]

Cool. Well done!