<PRE><B>Question 87086</B>
a)
The difference is the factor between each term. So going from 2 to 4, 4 to 6, 6 to 8, and 8 to 10 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 10: I&#39;m going to subtract 8 from it (which is the previous term) to get a difference of 2. If I pick 8, and subtract 6, I get a difference of 2. 


So the difference is: d=2




b)
Using what we found earlier, I know that the sequence counts up by 2 each term. So if I&#39;m at 2 (the 1st term) and I go to 4, this means I increase by 2 each term. If I let n=0 then the term is 2, and if I let n=1 then the term is 4. This basically tells me that the arithmetic sequence is 2n+2. To verify, simply plug in the 1st term (n=0) and you&#39;ll get 2. Plug in the 2nd term (n=1) you&#39;ll get 4, if I let n=2 I get 6, etc. If I wanted to know the 101st term, let n=100 (remember zero is the first term) and it comes to
{{{2*highlight(100)+2=202}}} So the 101st term is 202



c)Now lets find the sum of the first 20 terms


Using the sum of arithmetic series formula:
{{{s=(n/2)*(a[1]+a[n])}}} a[1]=first term, a[n]=nth term (ending term which is the 20th term), and n is the number of terms

Since we know the first term is 2, we know that {{{a[1]=2}}}. 

So lets calculate the 20th term. 


Let n=19(remember we started at n=0)


{{{a[19]=2(19)+2=38+2=40}}}


So the term {{{a[n]=40}}}


Now lets evaluate the sum

{{{s=(20/2)*(2+40)}}} Plug in values.  
{{{s=(10)*(42)}}}Simplify
{{{s=420}}} So the sum of the first 20 terms is 420.




d)Now lets find the sum of the first 30 terms


Using the sum of arithmetic series formula:
{{{s=(n/2)*(a[1]+a[n])}}} a[1]=first term, a[n]=nth term (ending term which is the 30th term), and n is the number of terms

Since we know the first term is 2, we know that {{{a[1]=2}}}. 

So lets calculate the 30th term. 


Let n=29(remember we started at n=0)


{{{a[19]=2(29)+2=38+2=60}}}


So the term {{{a[n]=60}}}


Now lets evaluate the sum

{{{s=(30/2)*(2+60)}}} Plug in values.  
{{{s=(15)*(62)}}}Simplify
{{{s=930}}} So the sum of the first 30 terms is 930.




e)


Sum of the first 2 terms 
2+4=6 
Sum of the first 3 terms 
2+4+6=12 
Sum of the first 4 terms 
2+4+6+8=20 
Sum of the first 5 terms 
2+4+6+8+10=30 
Sum of the first 6 terms 
2+4+6+8+10+12=42 
Sum of the first 7 terms 
2+4+6+8+10+12+14=56 
Sum of the first 8 terms 
2+4+6+8+10+12+14+16=72 
Sum of the first 9 terms 
2+4+6+8+10+12+14+16+18=90 
Sum of the first 10 terms 
2+4+6+8+10+12+14+16+18+20=110



Notice how the first  sum (which is 6) can be factored like this:
6=2*3

Notice how the second  sum (which is 12) can be factored like this:
12=3*4

Notice how the third  sum (which is 20) can be factored like this:
20=4*5

Notice how the fourth  sum (which is 30) can be factored like this:
30=5*6

Notice how the fifth  sum (which is 42) can be factored like this:
42=6*7

Notice how the sixth  sum (which is 56) can be factored like this:
56=7*8

Notice how the seventh  sum (which is 72) can be factored like this:
72=8*9

Notice how the eighth  sum (which is 90) can be factored like this:
90=9*10

Notice how the ninth  sum (which is 110) can be factored like this:
110=10*11



and this pattern continues...


Basically, every sum of this sequence can be factored into two numbers in which the second number is one more than the first number

So each sum follows the sequence {{{n(n+1)}}}   



Notice if we let {{{n=2}}} we get


{{{2(2+1)=2*3=6}}}


which is one of our sums. So this helps verify our sequence</PRE>