SOLUTION: The 15th term of an arithmetic series is 52, and the sum of the first 15 terms is 405. Find the first term of the series. This problem is from the Resource Book for McDougal Li

Algebra ->  Sequences-and-series -> SOLUTION: The 15th term of an arithmetic series is 52, and the sum of the first 15 terms is 405. Find the first term of the series. This problem is from the Resource Book for McDougal Li      Log On


   

Question 141060: The 15th term of an arithmetic series is 52, and the sum of the first 15 terms is 405. Find the first term of the series.
This problem is from the Resource Book for McDougal Littell's Algebra 2 book -- The worksheet is called :Challenge: Skills and Applications, Lesson 11.2
Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
The 15th term of an arithmetic series is 52, and the sum of the first 15 terms is 405. Find the first term of the series. 

Your book may use the letter "a" where I use the letter "t".
Some books use "t" and some "a".

Given: n=15, t15 = 52,  S15 = 405

You need two formulas.

1. The nth term of an arithmetic series is given by the formula

   tn = t1 + (n-1)d

2. The sum of the first n terms of an arithmetic series is given by
   the formula

   Sn = n%2F2[2t1 + (n-1)d]

Substituting in the first formula:

   t15 = t1 + (15-1)d
     52 = t1 + 14d


Substituting in the second formula:

   S15 = 15%2F2[2t1 + (15-1)d]

    405 = 15%2F2(2t1 + 14d)

Clear of fractions by multiplying both sides by 2

    810 = 15(2t1 + 14d)

Divide both sides by 15

     54 = 2t1 + 14d

So you have this system of equations:


     52 = t1 + 14d
     54 = 2t1 + 14d

Can you solve that system using substitution
or elimination?  If not post again asking how.

Answer:       t1 = 2. d = 25/7  

Edwin