SOLUTION: the 3rd and 7th term of an arithmetic progression are 18 and 30 respectively, the sum of the first 33 terms is?

Algebra ->  Sequences-and-series -> SOLUTION: the 3rd and 7th term of an arithmetic progression are 18 and 30 respectively, the sum of the first 33 terms is?       Log On


   

Question 1112814: the 3rd and 7th term of an arithmetic progression are 18 and 30 respectively, the sum of the first 33 terms is?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(1) Find the common difference, given 3rd term 18 and 7th term 30.
30+=+18%2B4d
12+=+4d
d+=+3

(2) Find the first term, knowing 3rd term 18 and common difference 3.
18-2d+=+18-6+=+12

(3) Find the 33rd term
12%2B32d+=+12%2B32%283%29+=+12%2B96+=+108

(4) Find the average of the first and 33rd terms, which is also the average of all the terms
%2812%2B108%29%2F2+=+120%2F2+=+60

(5) The sum of the first 33 terms is the number of terms, multiplied by the average of the terms
33%2A60+=+1980

Answer: The sum of the first 33 terms is 1980.



Note you can get to the answer a bit faster by noting that, in an arithmetic sequence of 33 terms, the 17th term is in the middle, and so it so is the average of all the terms. Then after the first two steps above you can do this:

(3) Find the 17th term
12%2B16d+=+12%2B16%283%29+=+12%2B48+=+60

(4) The sum of the first 33 terms is the number of terms, multiplied by the average of the terms
33%2A60+=+1980