College algebra

Question

How many terms of the arithmetic sequence 6,10,14,... must be added together to get 4606

Answer 1

The formula for the sum of the first n items in an Arithmetic Sequence is:

     SUM = (n/2)(2a + (n-1)d)        where a is first element; d is the constant difference

 

In this problem,

      a = 6

      d = 4

 

So. how many elements (that is, n) does it take to add up to 4606?

 

      4606 = (n/2)(2*6 + (n-1)*4)

      4606 = n*(6 + (n-1)*2)

      4606 = 6n + 2n² - 2n

      4606 = 2n² + 4n

       n² +2n - 2303 = 0

      (n-47)(n+49) = 0

   Either n=47  or  n=-49 (not possible)

 

   The sum of 47 terms is 4606.

            

Answer 2

4606 = (n/2)(2(4) + (n - 1)4)

4606 = (n/2)(8 + 4n - 4)

4606 = (n/2)(4 - 4n)

 

Can you solve for n and answer?