Sequences - geometric and arithmetic

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.

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 17

3 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.

Thanks Guv, this really helps :)

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

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)

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 :(.

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

Cool. Well done!