Question 69247
a)
The distance is the factor between each term. So going from 2 to 4, 4 to 6, 6 to 8, you see that its adding 2 each time. To verify pick one term and subtract the previous term from it. So lets say I choose 8: I'm going to subtract 6 from it to get a difference of 2. If I pick 6, and subtract 4, I get a difference of 2.


So the distance (d) is

d=2


b)
Using what we found earlier, I know that the sequence counts up by 2 each term. So if I'm at 8 (the 4th term) then my nth term is 4 (if we started at n=1). If I let n=5 then the term is 10. So if I divide 10 by 5 then I get 2, which means the term is doubled. This basically tells me that the arithmetic sequence is 2n. To verify, simply plug in the 1st term (n=1) and you'll get 2. Plug in the 2nd term (n=2) you'll get 4, etc. If I wanted to know the 101st term, let n=101 and it comes to
{{{2*highlight(101)=202}}} So the 101st term is 202

c)
Using the sum of arithmetic series formula:
{{{s=(n/2)*(a[1]+a[n])}}} a[1]=first term, a[n]=nth term, and n is nth term
{{{s=(20/2)*(2+40)}}} Plug in values
{{{s=(10)*(42)}}}Simplify
{{{s=420}}} So the sum of the first 20 terms is 420 (A minor discrepancy with your answer, I'd like to know what you did to get 430)

d)
Again using the same formula
{{{s=(n/2)*(a[1]+a[n])}}} a[1]=first term, a[n]=nth term, and n is nth term
{{{s=(30/2)*(2+60)}}} Plug in values
{{{s=(15)*(62)}}}Simplify
{{{s=930}}} close but no cigar, maybe you added the same term twice?

e)
This one is a little tricky to explain, but the series (partial sums) follow the sequence {{{n*(n+1)}}} (try it out by picking n=1,n=2,n=3,etc). To make a long story short, this sequence is the sum of n even integers.