SOLUTION: hi, I'm Sarah and i'm in 10th grade taking algebra 2, but i'm also taking an SAT prep course, and i'm having a little trouble with questions involving finding the nth term. it's qu

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Question 60451: hi, I'm Sarah and i'm in 10th grade taking algebra 2, but i'm also taking an SAT prep course, and i'm having a little trouble with questions involving finding the nth term. it's quite a bit more advanced than regular math, but i hope you can still give me a hand. here is an example of one of the problems i was given:
Instructions: writs the general term of the sequence:
1,8,27,64,125,...
(answer goes here) -> Nth term:
i was also given problems using variables; but i can't understand them without first understanding how to find the answer using numerical values. i would be most appreciative if you could also work one of the questions with variables so i can get an idea of how to work them. here is a question with variables:
The sequence a, a+d, a+2d, a+3d, a+4d,... has as it's Nth or general term (answer goes in blank) -> _________
thank you in advance for your help!

Found 2 solutions by hayek, josmiceli:
Answer by hayek(51) About Me  (Show Source):
You can put this solution on YOUR website!
1,8,27,64,125,...
You need to identify the relationship between the numbers. All of these numbers are perfect cubes, so the relationship is:

For the terms with variables, the relationship is already given to you:


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Look at the numbers in terms of what powers they are of what numbers
2%5E2+=+4
2%5E3+=+8
2%5E4+=+16 etc.
and
3%5E2+=+9
3%5E3+=+27
3%5E4+=+81 etc.
look at your sequence number by number
1,8,27,64,125
1+=+1%5E2
8+=+2%5E3
27+=+3%5E3
64+=+4%5E3
125+=+5%5E3
The first one is out of place but 1%5E2+=+1 and 1%5E3+=+1 also
so, the sequence is n%5E3 where n is the position in the sequence.
Test yourself. What is the 11th term?
1331+=+11%5E3
what is the 10,000th term?
10%5E12+=+10000%5E3
------------------
a, a+d, a+2d, a+3d, a+4d,...
a is the same in each term so, it's a + something
The multiplier for d looks like its position in the series minus one
n = position
a+%2B+%28n-1%29d answer
test this to see if it works for each term
1st term: a+%2B+%281-1%29d+=+a
2nd term: a+%2B+%282-1%29d+=+a+%2B+d etc.