SOLUTION: PLEASE HELP ME I AM STUCK.....
AN ARITHMETIC PROGRESSION SERIES HAS THE FIRST TERM 4 AND THE COMMON DIFFERENCE OF 3 .IF THE SUM OF ALL INTEGERS IS 375.HOW MANY TERMS ARE THERE?
P
Algebra ->
Sequences-and-series
-> SOLUTION: PLEASE HELP ME I AM STUCK.....
AN ARITHMETIC PROGRESSION SERIES HAS THE FIRST TERM 4 AND THE COMMON DIFFERENCE OF 3 .IF THE SUM OF ALL INTEGERS IS 375.HOW MANY TERMS ARE THERE?
P
Log On
Question 936727: PLEASE HELP ME I AM STUCK.....
AN ARITHMETIC PROGRESSION SERIES HAS THE FIRST TERM 4 AND THE COMMON DIFFERENCE OF 3 .IF THE SUM OF ALL INTEGERS IS 375.HOW MANY TERMS ARE THERE?
PLEASE SOLVE IT AND EXPLAIN ME TOO Found 2 solutions by Danielshinmath, MathTherapy:Answer by Danielshinmath(31) (Show Source):
You can put this solution on YOUR website! The sum is 4n+3*(n-1)=4n+3n-3=375, where n is the amount of terms and 4 represents thee first term. The 3 represents the common difference.
if 4n+3n-3=375, that means 7n=375+3, which equals 7n=378. Then you get the answer n-378/7, or 54.
