If the 13th and 38th terms of an arithmetic sequence are -53 and -128, respectively, find the 668th term of the
series.

9514 1404 393

  -2018

Step-by-step explanation:

The n-th term is ...

  an = a1 +d(n -1)

So, the given terms are ...

  -53 = a1 +12d

  -128 = a1 +37d

Subtracting the first from the second gives ...

  (a1 +37d) -(a1 +12d) = (-128) -(-53)

  25d = -75

  d = -3

The 668th term will be ...

  a668 = a1 +d(668 -1) = a1 +667d = (a1 +37d) +630d

  a668 = -128 +630(-3) = -128 -1890 . . . . substitute for a38

  a668 = -2018